167
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MAT2305 LINEAR AL :Week4 Thanatyod Jampawai, Ph.D. Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 1 / 40

# · ........................................ Determinant Prperties of thr Determinant Solution det(A) = 1, det(B) = 10 and det(C) = 9 1. det(AB) = det(A)det(B)= ( 1)( 10

others

• View
17

0

Embed Size (px)

Citation preview

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

MAT2305 LINEAR AL :Week4

Thanatyod Jampawai, Ph.D.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 1 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Review Week 3

Diagonal MatrixSymmetric Matrix(Upper and Lower) Triangular MatrixLU DecompositionPermutation and InversionEven and OddElementary ProductSigned Elementary ProductDeterminat Function

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 2 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Outline for Week 3

Chapter 3 Determinant3.3 Properties of the Determinant3.4 Adjoint of a Matrix3.5 Cramer’s Rule

Assignment 4

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 3 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

3.3 Properties of the Determinant

Let A,B and C be n× n matrices that differ only in a single row, say the rth,and assume that the rth row of C can be obtained by adding correspondingentries in the rth rows of A and B. Then

det(C) = det(A) + det(B)

∣∣∣∣a bc d

∣∣∣∣+ ∣∣∣∣a bx y

∣∣∣∣ = (ad− bc) + (ay − bx) = a(d+ y)− b(c+ x) =

∣∣∣∣ a bc+ x d+ y

∣∣∣∣

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 4 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

3.3 Properties of the Determinant

Let A,B and C be n× n matrices that differ only in a single row, say the rth,and assume that the rth row of C can be obtained by adding correspondingentries in the rth rows of A and B. Then

det(C) = det(A) + det(B)

∣∣∣∣a bc d

∣∣∣∣+ ∣∣∣∣a bx y

∣∣∣∣ = (ad− bc) + (ay − bx) = a(d+ y)− b(c+ x) =

∣∣∣∣ a bc+ x d+ y

∣∣∣∣

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 4 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.1)

By evaluating the determinant, check

det

1 7 52 0 3

1 + 0 4 + 1 7 + (−1)

= det

1 7 52 0 31 4 7

+ det

1 7 52 0 30 1 −1

Solution

∣∣∣∣∣∣1 7 52 0 3

1 + 0 4 + 1 7 + (−1)

∣∣∣∣∣∣ =∣∣∣∣∣∣1 7 52 0 31 5 6

∣∣∣∣∣∣ = −28

∣∣∣∣∣∣1 7 52 0 31 4 7

∣∣∣∣∣∣+∣∣∣∣∣∣1 7 52 0 30 1 −1

∣∣∣∣∣∣ = −49 + 21 = −28

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 5 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.1)

By evaluating the determinant, check

det

1 7 52 0 3

1 + 0 4 + 1 7 + (−1)

= det

1 7 52 0 31 4 7

+ det

1 7 52 0 30 1 −1

Solution

∣∣∣∣∣∣1 7 52 0 3

1 + 0 4 + 1 7 + (−1)

∣∣∣∣∣∣ =∣∣∣∣∣∣1 7 52 0 31 5 6

∣∣∣∣∣∣ = −28

∣∣∣∣∣∣1 7 52 0 31 4 7

∣∣∣∣∣∣+∣∣∣∣∣∣1 7 52 0 30 1 −1

∣∣∣∣∣∣ = −49 + 21 = −28

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 5 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Scalar Multiplication to Determinant

Theorem (3.3.2)

Let A be an n× n matrix and k be any scalar. Then

det(kA) = kn det(A)

Example (3.3.2)

Evaluate the determinant of the matrix

det(2A), det(−A), det(3AT ) and det(−(−2A)T )

if det(A) = −5 for 3× 3 matrix A.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 6 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Scalar Multiplication to Determinant

Theorem (3.3.2)

Let A be an n× n matrix and k be any scalar. Then

det(kA) = kn det(A)

Example (3.3.2)

Evaluate the determinant of the matrix

det(2A), det(−A), det(3AT ) and det(−(−2A)T )

if det(A) = −5 for 3× 3 matrix A.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 6 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A)

= 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A)

= 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A)

= det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A)

= (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)

= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT )

= 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )

= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)

= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T )

= (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )

= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )

= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)

= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(2A) = 23 det(A) = 8(−5) = −40 #

det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #

det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #

det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )

= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Invertible Test

Theorem (3.3.3)

If B is an n× n matrix and E is n× n elementary matrix, then

det(EB) = det(E) det(B)

Moreover, if E1, E2, ..., Er are n× n elementary matrices, then

det(E1E2 · · ·ErB) = det(E1) det(E2) · · · det(Er) det(B).

Theorem (3.3.4 Invertible Test)

A square matrix A is invertible if and only if det(A) ̸= 0 .

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 8 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Invertible Test

Theorem (3.3.3)

If B is an n× n matrix and E is n× n elementary matrix, then

det(EB) = det(E) det(B)

Moreover, if E1, E2, ..., Er are n× n elementary matrices, then

det(E1E2 · · ·ErB) = det(E1) det(E2) · · · det(Er) det(B).

Theorem (3.3.4 Invertible Test)

A square matrix A is invertible if and only if det(A) ̸= 0 .

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 8 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.3)

Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.

1. A =

1 0 10 2 40 0 4

det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)

2. A =

1 2 30 −5 72 4 6

det(A) = 0 (NOT invertible)

3. A =

0 1 11 0 00 1 0

det(A) = 1 ̸= 0 (Invertible)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.3)

Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.

1. A =

1 0 10 2 40 0 4

det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)

2. A =

1 2 30 −5 72 4 6

det(A) = 0 (NOT invertible)

3. A =

0 1 11 0 00 1 0

det(A) = 1 ̸= 0 (Invertible)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.3)

Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.

1. A =

1 0 10 2 40 0 4

det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)

2. A =

1 2 30 −5 72 4 6

det(A) = 0 (NOT invertible)

3. A =

0 1 11 0 00 1 0

det(A) = 1 ̸= 0 (Invertible)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.3)

Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.

1. A =

1 0 10 2 40 0 4

det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)

2. A =

1 2 30 −5 72 4 6

det(A) = 0 (NOT invertible)

3. A =

0 1 11 0 00 1 0

det(A) = 1 ̸= 0 (Invertible)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.3)

Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.

1. A =

1 0 10 2 40 0 4

det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)

2. A =

1 2 30 −5 72 4 6

det(A) = 0 (NOT invertible)

3. A =

0 1 11 0 00 1 0

det(A) = 1 ̸= 0 (Invertible)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.3)

Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.

1. A =

1 0 10 2 40 0 4

det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)

2. A =

1 2 30 −5 72 4 6

det(A) = 0 (NOT invertible)

3. A =

0 1 11 0 00 1 0

det(A) = 1 ̸= 0 (Invertible)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Product Determinant

Theorem (3.3.5)

If A and B are square matrix of the same size, then

det(AB) = det(A) det(B)

If m is a positive integer,

det(Am) = [det(A)]m

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 10 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Example (3.3.4)

Let A =

[1 23 5

], B =

[−2 26 −1

]and C =

[11 45 1

]. Find

1. det(AB)

2. det(BC)

3. det(ATC)

4. det(2AB)

5. det(−3ABTC)

6. det(2C(AB2)T )

7. det(A2B3C2)

8. det(A+AT )

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 11 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB)

= det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)

= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC)

= det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)

= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC)

= det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)

= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)

= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB)

= 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)

= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)

= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC)

= (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)

= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T )

= 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)

= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2)

= [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2

= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]

det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −1, det(B) = −10 and det(C) = −9

1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #

3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #

4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #

5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #

6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #

7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #

8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Determinant of Inverse

Theorem (3.3.6)

If A is invertible, then

det(A−1) =1

det(A)

Example (3.3.6)

Let A =

[6 −4−3 1

]and B =

[8 45 1

]. Find

1. det(A−1)

2. det(AB−1)

3. det(3(−2AT )−1)

4. det(2(−3A)TA−1BT )

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 13 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Determinant of Inverse

Theorem (3.3.6)

If A is invertible, then

det(A−1) =1

det(A)

Example (3.3.6)

Let A =

[6 −4−3 1

]and B =

[8 45 1

]. Find

1. det(A−1)

2. det(AB−1)

3. det(3(−2AT )−1)

4. det(2(−3A)TA−1BT )

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 13 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1)

= 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1)

= det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)

= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1)

= 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)

= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )

= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Solution

det(A) = −6 and det(B) = −12

1. det(A−1) = 1det(A)

= − 16

#

2. det(AB−1) = det(A) det(B−1)= (−6) 1−12

= 12

#

3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )

= 9(−2)2 det(A)

= 94(−6)

= − 38

#

4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )

= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Prperties of thr Determinant

Equivalent Statements

Theorem (3.3.7)

Let A be an n× n matrix. Then TFAE.

1 A is invertible.

2 Ax = 0 has only trivial solution.

3 The RREF of A is In.

4 A is expressible as a product of elementary matrices.

5 Ax = b is consistent for every n× 1 matrix b.

6 Ax = b has exactly one solution for every n× 1 matrix b.

7 det(A) ̸= 0.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 15 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Definition (3.4.1 Minor and Cofactor)

Let A be a square matrix.

The minor of entry aij is denoted by Mij and is defined to be thedeterminant of the submatrix that remains after the ith row and jthcolumn are deleted from A.

The number (−1)i+jMij is denoted by Cij and is called the cofactor ofentry aij,

Cij = (−1)i+jMij

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 16 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.1)

Find all minors and all cofactors of matrix

A =

1 2 12 1 11 2 0

Solution

M11 =

∣∣∣∣1 12 0

∣∣∣∣ = −2

M12 =

∣∣∣∣2 11 0

∣∣∣∣ = −1

M13 =

∣∣∣∣2 11 2

∣∣∣∣ = 3

M21 =

∣∣∣∣2 12 0

∣∣∣∣ = −2

M22 =

∣∣∣∣1 11 0

∣∣∣∣ = −1

M23 =

∣∣∣∣1 21 2

∣∣∣∣ = 0

M31 =

∣∣∣∣2 11 1

∣∣∣∣ = 1

M32 =

∣∣∣∣1 12 1

∣∣∣∣ = −1

M33 =

∣∣∣∣1 22 1

∣∣∣∣ = −3

Thus,

C11 = (−1)1+1M11 = −2

C12 = (−1)1+2M12 = 1

C13 = (−1)1+3M13 = 3

C21 = (−1)2+1M21 = 2

C22 = (−1)2+2M22 = −1

C23 = (−1)2+3M23 = 0

C31 = (−1)3+1M31 = 1

C32 = (−1)3+2M32 = 1

C33 = (−1)3+3M33 = −3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Expansions by Cofactors

Theorem (3.4.1)

The determinant of an n× n matrix A = [aij ] can be computed by

cofactor expansion along the jth column:

det(A) = a1jC1j + a2jC2j + · · ·+ anjCnj

cofactor expansion along the ith row:

det(A) = ai1Ci1 + ai2Ci2 + · · ·+ ainCin

for each 1 ≤ i ≤ n and 1 ≤ j ≤ n.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 18 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row

det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13

= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column

det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31

= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row

det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33

= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.2)

Find det(A) in example 3.4.1 by cofactor expansions.

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Thus,

First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3

First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3

Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.3)

Find determinant of the following matrices by row operations or cofactorexpansions or both.

1.

1 3 50 2 20 3 −1

2.

3 5 1 50 0 −1 12 4 0 5−1 0 9 0

3.

3 0 0 01 2 −1 12 4 0 53 4 5 3

4.

3 5 −2 61 2 −1 12 4 1 53 7 5 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 20 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

1. Cofactor expansion

∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1

∣∣∣∣2 23 −1

∣∣∣∣ = −8 #

Row Operation∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 0 −4

∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

1. Cofactor expansion∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1

∣∣∣∣2 23 −1

∣∣∣∣ = −8 #

Row Operation∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 0 −4

∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

1. Cofactor expansion∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1

∣∣∣∣2 23 −1

∣∣∣∣ = −8 #

Row Operation

∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 0 −4

∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

1. Cofactor expansion∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1

∣∣∣∣2 23 −1

∣∣∣∣ = −8 #

Row Operation∣∣∣∣∣∣1 3 50 2 20 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 3 −1

∣∣∣∣∣∣ = 2

∣∣∣∣∣∣1 3 50 1 10 0 −4

∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

2. Cofactor expansion

∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24

= −1 · (−1)2+3

∣∣∣∣∣∣3 5 52 4 5−1 0 0

∣∣∣∣∣∣+ 1 · (−1)2+4

∣∣∣∣∣∣3 5 12 4 0−1 0 9

∣∣∣∣∣∣= −5 + 22 = 17 #

Row Operation∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5

∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1

∣∣∣∣∣∣0 −1 14 18 55 28 5

∣∣∣∣∣∣= 17 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

2. Cofactor expansion∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24

= −1 · (−1)2+3

∣∣∣∣∣∣3 5 52 4 5−1 0 0

∣∣∣∣∣∣+ 1 · (−1)2+4

∣∣∣∣∣∣3 5 12 4 0−1 0 9

∣∣∣∣∣∣= −5 + 22 = 17 #

Row Operation∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5

∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1

∣∣∣∣∣∣0 −1 14 18 55 28 5

∣∣∣∣∣∣= 17 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

2. Cofactor expansion∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24

= −1 · (−1)2+3

∣∣∣∣∣∣3 5 52 4 5−1 0 0

∣∣∣∣∣∣+ 1 · (−1)2+4

∣∣∣∣∣∣3 5 12 4 0−1 0 9

∣∣∣∣∣∣= −5 + 22 = 17 #

Row Operation

∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5

∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1

∣∣∣∣∣∣0 −1 14 18 55 28 5

∣∣∣∣∣∣= 17 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

2. Cofactor expansion∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24

= −1 · (−1)2+3

∣∣∣∣∣∣3 5 52 4 5−1 0 0

∣∣∣∣∣∣+ 1 · (−1)2+4

∣∣∣∣∣∣3 5 12 4 0−1 0 9

∣∣∣∣∣∣= −5 + 22 = 17 #

Row Operation∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5

∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1

∣∣∣∣∣∣0 −1 14 18 55 28 5

∣∣∣∣∣∣= 17 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

3. Cofactor expansion

∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = 3C11 = 3

∣∣∣∣∣∣2 −1 14 0 54 5 3

∣∣∣∣∣∣ = 3(−38) = −114 #

Row Operation∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣ = −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5

∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

3. Cofactor expansion∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = 3C11 = 3

∣∣∣∣∣∣2 −1 14 0 54 5 3

∣∣∣∣∣∣ = 3(−38) = −114 #

Row Operation∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣ = −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5

∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

3. Cofactor expansion∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = 3C11 = 3

∣∣∣∣∣∣2 −1 14 0 54 5 3

∣∣∣∣∣∣ = 3(−38) = −114 #

Row Operation

∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣ = −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5

∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

3. Cofactor expansion∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = 3C11 = 3

∣∣∣∣∣∣2 −1 14 0 54 5 3

∣∣∣∣∣∣ = 3(−38) = −114 #

Row Operation∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0

∣∣∣∣∣∣∣∣ = −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1

∣∣∣∣∣∣∣∣= −3

∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5

∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

4.

3 5 −2 61 2 −1 12 4 1 53 7 5 3

Row Operation∣∣∣∣∣∣∣∣

3 5 −2 61 2 −1 12 4 1 53 7 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 13 5 −2 62 4 1 53 7 5 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 10 −1 1 30 0 3 30 1 8 0

∣∣∣∣∣∣∣∣= −

∣∣∣∣∣∣∣∣1 2 −1 10 −1 1 30 0 3 30 0 9 3

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 2 −1 10 −1 1 30 0 3 30 0 0 −6

∣∣∣∣∣∣∣∣= (−1)(1)(−1)(3)(−6) = −18 #

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 24 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Take a BreakFor 10 Minutes

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 25 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Definition (3.4.2)

Let A be any n× n matrix and Cij be the cofactor of aij. The the matrixC11 C12 · · · C1n

C21 C22 · · · C2n

......

...Cn1 Cn2 · · · Cnn

is called the matrix of cofactor from A. The transpose of this matrix is

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 26 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.4)

Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Co(A) =

−2 1 32 −1 01 1 −3

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.4)

Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Co(A) =

−2 1 32 −1 01 1 −3

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.4)

Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Co(A) =

−2 1 32 −1 01 1 −3

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.4)

Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

C11 = −2

C12 = 1

C13 = 3

C21 = 2

C22 = −1

C23 = 0

C31 = 1

C32 = 1

C33 = −3

Co(A) =

−2 1 32 −1 01 1 −3

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Inverse of Matrix Using its Adjoint

Theorem (3.4.2)

If A is invertible matrix, then

A−1 =1

Example (3.4.5)

Use adjoint to find the inverse of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

A−1 = 1det(A)

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 28 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Inverse of Matrix Using its Adjoint

Theorem (3.4.2)

If A is invertible matrix, then

A−1 =1

Example (3.4.5)

Use adjoint to find the inverse of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

A−1 = 1det(A)

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 28 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Inverse of Matrix Using its Adjoint

Theorem (3.4.2)

If A is invertible matrix, then

A−1 =1

Example (3.4.5)

Use adjoint to find the inverse of the matrix A in example 3.4.1

A =

1 2 12 1 11 2 0

Solution

A−1 = 1det(A)

−2 2 11 −1 13 0 −3

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 28 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.6)

Use adjoint to find the inverse of the matrix A where

A =

1 3 42 4 26 3 −1

Solution

Then det(A) = −40. Thus,

A−1

=1

1

40

C11 C21 C31C12 C22 C32C13 C23 C33

= −1

40

+

∣∣∣∣4 23 −1

∣∣∣∣ −∣∣∣∣3 43 −1

∣∣∣∣ +

∣∣∣∣3 44 2

∣∣∣∣−

∣∣∣∣2 26 −1

∣∣∣∣ +

∣∣∣∣1 46 −1

∣∣∣∣ −∣∣∣∣1 42 2

∣∣∣∣+

∣∣∣∣2 46 3

∣∣∣∣ −∣∣∣∣1 36 3

∣∣∣∣ +

∣∣∣∣1 32 4

∣∣∣∣

= −1

40

−10 15 −1014 −25 6−18 15 −2

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 29 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.6)

Use adjoint to find the inverse of the matrix A where

A =

1 3 42 4 26 3 −1

Solution

Then det(A) = −40.

Thus,

A−1

=1

1

40

C11 C21 C31C12 C22 C32C13 C23 C33

= −1

40

+

∣∣∣∣4 23 −1

∣∣∣∣ −∣∣∣∣3 43 −1

∣∣∣∣ +

∣∣∣∣3 44 2

∣∣∣∣−

∣∣∣∣2 26 −1

∣∣∣∣ +

∣∣∣∣1 46 −1

∣∣∣∣ −∣∣∣∣1 42 2

∣∣∣∣+

∣∣∣∣2 46 3

∣∣∣∣ −∣∣∣∣1 36 3

∣∣∣∣ +

∣∣∣∣1 32 4

∣∣∣∣

= −1

40

−10 15 −1014 −25 6−18 15 −2

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 29 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.6)

Use adjoint to find the inverse of the matrix A where

A =

1 3 42 4 26 3 −1

Solution

Then det(A) = −40. Thus,

A−1

=1

1

40

C11 C21 C31C12 C22 C32C13 C23 C33

= −1

40

+

∣∣∣∣4 23 −1

∣∣∣∣ −∣∣∣∣3 43 −1

∣∣∣∣ +

∣∣∣∣3 44 2

∣∣∣∣−

∣∣∣∣2 26 −1

∣∣∣∣ +

∣∣∣∣1 46 −1

∣∣∣∣ −∣∣∣∣1 42 2

∣∣∣∣+

∣∣∣∣2 46 3

∣∣∣∣ −∣∣∣∣1 36 3

∣∣∣∣ +

∣∣∣∣1 32 4

∣∣∣∣

= −1

40

−10 15 −1014 −25 6−18 15 −2

#

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 29 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Entry of Inverse Matrix

Theorem (3.4.3)

Let A = [aij ] be an n× n matrix and Cij be the cofactor of aij. If A isinvertible and A−1 = [a∗ij ], then

a∗ij =1

det(A)Cji

[a∗ij ] = A−1 =1

det(A)

C11 C21 · · · Cj1 · · · Cn1

C12 C22 · · · Cj2 · · · Cn2

...C1i C2i · · · Cji · · · Cni

...C1n C2n · · · Cjn · · · Cnn

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 30 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Entry of Inverse Matrix

Theorem (3.4.3)

Let A = [aij ] be an n× n matrix and Cij be the cofactor of aij. If A isinvertible and A−1 = [a∗ij ], then

a∗ij =1

det(A)Cji

[a∗ij ] = A−1 =1

det(A)

C11 C21 · · · Cj1 · · · Cn1

C12 C22 · · · Cj2 · · · Cn2

...C1i C2i · · · Cji · · · Cni

...C1n C2n · · · Cjn · · · Cnn

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 30 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.7)

Use theorem 3.4.3 to compute entry of the inverse of matrix A where

A =

1 3 4 12 4 2 00 2 1 01 3 −1 0

If A−1 = [a∗ij ], find a∗12 and a∗42

Solution

det(A) = 1 · C14 = −

∣∣∣∣∣∣2 4 20 2 11 3 −1

∣∣∣∣∣∣ =∣∣∣∣∣∣1 3 −10 2 12 4 2

∣∣∣∣∣∣R13 =

∣∣∣∣∣∣1 3 −10 2 10 −2 4

∣∣∣∣∣∣R3 − 2R1

=

∣∣∣∣∣∣1 3 −10 2 10 0 5

∣∣∣∣∣∣R3 +R2

= (1)(2)(5) = 10

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 31 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Example (3.4.7)

Use theorem 3.4.3 to compute entry of the inverse of matrix A where

A =

1 3 4 12 4 2 00 2 1 01 3 −1 0

If A−1 = [a∗ij ], find a∗12 and a∗42

Solution

det(A) = 1 · C14 = −

∣∣∣∣∣∣2 4 20 2 11 3 −1

∣∣∣∣∣∣ =∣∣∣∣∣∣1 3 −10 2 12 4 2

∣∣∣∣∣∣R13 =

∣∣∣∣∣∣1 3 −10 2 10 −2 4

∣∣∣∣∣∣R3 − 2R1

=

∣∣∣∣∣∣1 3 −10 2 10 0 5

∣∣∣∣∣∣R3 +R2

= (1)(2)(5) = 10

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 31 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

a∗12 =1

det(A)C21 =

1

10(−1)2+1

∣∣∣∣∣∣3 4 12 1 03 −1 0

∣∣∣∣∣∣= −

1

10· 1 · (−1)1+3

∣∣∣∣2 13 −1

∣∣∣∣ = 5

10=

1

2#

a∗42 =1

det(A)C24 =

1

10(−1)2+4

∣∣∣∣∣∣1 3 40 2 11 3 −1

∣∣∣∣∣∣ = 1

10

∣∣∣∣∣∣1 3 40 2 10 0 −5

∣∣∣∣∣∣R3 −R1

=1

10(1)(2)(−5) = −1 #

Computed by www.wolframalpha.com A−1 = 110

0 5 −10 00 −1 4 20 2 2 −410 −10 −10 10

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 32 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

a∗12 =1

det(A)C21 =

1

10(−1)2+1

∣∣∣∣∣∣3 4 12 1 03 −1 0

∣∣∣∣∣∣= −

1

10· 1 · (−1)1+3

∣∣∣∣2 13 −1

∣∣∣∣ = 5

10=

1

2#

a∗42 =1

det(A)C24 =

1

10(−1)2+4

∣∣∣∣∣∣1 3 40 2 11 3 −1

∣∣∣∣∣∣ = 1

10

∣∣∣∣∣∣1 3 40 2 10 0 −5

∣∣∣∣∣∣R3 −R1

=1

10(1)(2)(−5) = −1 #

Computed by www.wolframalpha.com A−1 = 110

0 5 −10 00 −1 4 20 2 2 −410 −10 −10 10

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 32 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Solution

a∗12 =1

det(A)C21 =

1

10(−1)2+1

∣∣∣∣∣∣3 4 12 1 03 −1 0

∣∣∣∣∣∣= −

1

10· 1 · (−1)1+3

∣∣∣∣2 13 −1

∣∣∣∣ = 5

10=

1

2#

a∗42 =1

det(A)C24 =

1

10(−1)2+4

∣∣∣∣∣∣1 3 40 2 11 3 −1

∣∣∣∣∣∣ = 1

10

∣∣∣∣∣∣1 3 40 2 10 0 −5

∣∣∣∣∣∣R3 −R1

=1

10(1)(2)(−5) = −1 #

Computed by www.wolframalpha.com A−1 = 110

0 5 −10 00 −1 4 20 2 2 −410 −10 −10 10

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 32 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

3.5 Cramer’s Rule

Gabriel Cramer (31 July 1704 to 4 January 1752)

He was a Swiss mathematician, born in Geneva. He was the son of physician Jean Cramerand Anne Mallet Cramer.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 33 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Theorem (3.5.1)

Let a system of n equations in n unknowns be

Ax = b

with det(A) ̸= 0. Then the system has a unique solution and this solution is

x1 = det(A1)det(A) , x2 = det(A2)

det(A) , ..., xn = det(An)det(A)

where Aj is the matrix obtained by replacing the entrices in the jth column ofA by the entries in the matrix b.

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 34 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule

Ax = b → x = A−1b

x = A−1b =1

1

det(A)

C11 C21 · · · Cj1 · · · Cn1

C12 C22 · · · Cj2 · · · Cn2

...C1i C2i · · · Cji · · · Cni

...C1n C2n · · · Cjn · · · Cnn

b1b2...bi...bn

x =

x1

x2

...xi

...xn

=

1

det(A)

b1C11 + b2C21 + · · ·+ biCj1 + · · ·+ bnCn1

b1C12 + b2C22 + · · ·+ biCj2 + · · ·+ bnCn2

...b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

...b1C1n + b2C2n + · · · biCjn + · · ·+ bnCnn

xi =

b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 35 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule

Ax = b → x = A−1b

x = A−1b =1

1

det(A)

C11 C21 · · · Cj1 · · · Cn1

C12 C22 · · · Cj2 · · · Cn2

...C1i C2i · · · Cji · · · Cni

...C1n C2n · · · Cjn · · · Cnn

b1b2...bi...bn

x =

x1

x2

...xi

...xn

=

1

det(A)

b1C11 + b2C21 + · · ·+ biCj1 + · · ·+ bnCn1

b1C12 + b2C22 + · · ·+ biCj2 + · · ·+ bnCn2

...b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

...b1C1n + b2C2n + · · · biCjn + · · ·+ bnCnn

xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 35 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule

Ax = b → x = A−1b

x = A−1b =1

1

det(A)

C11 C21 · · · Cj1 · · · Cn1

C12 C22 · · · Cj2 · · · Cn2

...C1i C2i · · · Cji · · · Cni

...C1n C2n · · · Cjn · · · Cnn

b1b2...bi...bn

x =

x1

x2

...xi

...xn

=

1

det(A)

b1C11 + b2C21 + · · ·+ biCj1 + · · ·+ bnCn1

b1C12 + b2C22 + · · ·+ biCj2 + · · ·+ bnCn2

...b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

...b1C1n + b2C2n + · · · biCjn + · · ·+ bnCnn

xi =

b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 35 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule (Continue)

xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Set up,

Ai =

a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n

...an1 an2 · · · ani−1 bn ani+1 · · · ann

So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,

xi =det(Ai)

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule (Continue)

xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Set up,

Ai =

a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n

...an1 an2 · · · ani−1 bn ani+1 · · · ann

So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,

xi =det(Ai)

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule (Continue)

xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Set up,

Ai =

a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n

...an1 an2 · · · ani−1 bn ani+1 · · · ann

So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.

Therefore,xi =

det(Ai)

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Sketch Proof for Cramer’s Rule (Continue)

xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)

Set up,

Ai =

a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n

...an1 an2 · · · ani−1 bn ani+1 · · · ann

So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,

xi =det(Ai)

det(A)

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Example (3.5.1)

Use Cramer’s rule to solve

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 37 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣

=−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣

=0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣

=−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

1.

x1 + x3 = 4

x1 + x2 − x3 = −2

2x1 − x2 + x3 = 5

1 0 11 1 −12 −1 1

x1

x2

x3

=

4−25

x1 =

∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−3

−3= 1

x2 =

∣∣∣∣∣∣∣1 4 11 −2 −12 5 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

0

−3= 0

x3 =

∣∣∣∣∣∣∣1 0 41 1 −22 −1 5

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1

∣∣∣∣∣∣∣=

−9

−3= 3

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣

=18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣

=17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣

=−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Solution

2.

x+ y + 2z = 1

2x− y − 3z = −8

x− y + z = 2

1 1 22 −1 −31 −1 1

xyz

=

1−82

x1 =

∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

18

−11= −

18

11

x2 =

∣∣∣∣∣∣∣1 1 22 −8 −31 2 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

17

−11= −

17

11

x3 =

∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1

∣∣∣∣∣∣∣=

−23

−11=

23

11

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Determinant Cramer’s Rule

Assignment 4

1. Find determinant by EROs and Cofactor expansion.

(a) (EVEN)

2 2 2 −24 3 −4 1−6 −6 −16 01 1 3 5

(b) (ODD)

3 5 5 11 2 3 2−2 −5 −9 01 0 2 −1

2. (EVEN and ODD) Let A =

[1 3−2 −8

]2and B−1 =

[6 75 6

]. Use determinant properties

to evaluate their determinants.

(a) det(A)

(b) det(B)

(c) det(2A3B)

(d) det(3BTA−1)

(e) det((2A)−1BAT )

(f) det(AB +ABT )

3. Use Cramer’s rule to solve

(a) (EVEN)

x1 + x2 + x3 = 6

2x1 − x2 + x3 = 3

x1 + 3x2 − 2x3 = 1

(b) (ODD)

x1 − x2 + 3x3 = 10

x1 + x2 + 2x3 = 2

2x1 + 4x2 + x3 = −8

Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 40 / 40