6.5 determinant x

Determinant

The determinant of an nxn square matrix A,

or det(A), is a number.

Determinant



We define the determinant of a 1x1 matrix k ,

or det k to be k. Hence det –3 = –3.

Determinant





We define the determinant of the 2x2 matrix

or det

a b

c d a b

c d

Determinant






or det = ad

a b

c d a b

c d

Determinant






or det = ad – bc

a b

c d a b

c d

Determinant






or det = ad – bc

a b

c d a b

c d

Here are two motivations for this definition.

Determinant






or det = ad – bc

a b

c d a b

c d


I. (Geometric) Let (a, b) and (c, d)

be two points in the rectangular

coordinate system.

(a, b) (c, d)

Determinant






or det = ad – bc

a b

c d a b

c d


I. (Geometric) Let (a, b) and (c, d)

be two points in the rectangular

coordinate system.

Then deta b

c d = ad – bc

= “Signed” area of the parallelogram as shown.

(a, b) (c, d)

“Signed” area

= ad – bc

Determinant

DeterminantLet’s verify this in the simple cases where one of the

points is on the axis.



For example, Iet the points be

(a, 0) and (c, d) as shown.

(a, 0)

(c, d)





then the area of the

parallelogram is

deta 0

c d

(a, 0)

(c, d)

d

= ad






parallelogram is

deta 0

c d

(a, 0)

(c, d)

d

If we switch the order of the two rows then

detc d

= – ad

= ad

a 0






parallelogram is

deta 0

c d

(a, 0)

(c, d)

d


deta 0

c d = – ad

= ad

What’s the meaning of the “–” sign?

(besides that the rows are switched)






parallelogram is

deta 0

c d

(a, 0)

(c, d)

d


deta 0

c d = – ad

= ad

What’s the meaning of the “–” sign?

(besides that the rows are switched). Actually it tells

us the “relative position” of these two points.

Determinant

Given a point A(a, b) ≠ (0, 0), the dial with the tip at A

defines a direction.

A(a, b) A(a, b)

Determinant


defines a direction. Let B(c, d) be another point,

there are two possibilities, A(a, b) A(a, b)

Determinant



there are two possibilities,

either B is to the left or

it’s to the right of dial A

as shown.

A(a, b)

B(c, d)

A(a, b)

B(c, d)

Determinant






as shown. The determinant

identifies which case it is.

A(a, b)

B(c, d)

A(a, b)

B(c, d)

Determinant

If deta b

c d is +, then B is to the left of A.






as shown.

i.

The determinant

identifies which case it is. detAB

A(a, b)

B(c, d)

A(a, b)

B(c, d)

> 0

B is to the left of A.

Determinant

If deta b







as shown.

i.

ii.

The determinant


A(a, b)

B(c, d)

A(a, b)

B(c, d)

> 0 detAB < 0

B is to the left of A. B is to the right of A.

If deta b

c d is –, then B is to the right of A.

Determinant

If deta b







as shown.

i.

ii.

The determinant


A(a, b)

B(c, d)

A(a, b)

B(c, d)

> 0 detAB < 0

B is to the left of A. B is to the right of A.

If deta b

c d is –, then B is to the right of A.

For example det1 0

0 1 > 0 and det

1 0

0 1 < 0.

It is in the above sense that we have the

signed-area-answer of det A.

Determinant

So given the picture to the right: (a, b) (c, d)



Determinant

So given the picture to the right:

deta b

c d = + Area of

(a, b) (c, d)



Determinant


deta b

c d = + Area of

(a, b) (c, d)

deta b

c d = – Area of

and



Determinant


deta b

c d = + Area of

(a, b) (c, d)

deta b

c d = – Area of

and



Example A. Find the area.

Determinant

(5,–3)

(1,4)


deta b

c d = + Area of

(a, b) (c, d)

deta b

c d = – Area of

and




Determinant

(5,–3)

(1,4)

det5 –3

1 4

To find the area, we calculate


deta b

c d = + Area of

(a, b) (c, d)

deta b

c d = – Area of

and




Determinant

(5,–3)

(1,4)

det5 –3

1 4


= 5(4) – 1(–3) = 23 = area


deta b

c d = + Area of

(a, b) (c, d)

deta b

c d = – Area of

and




Determinant

(5,–3)

(1,4)

det5 –3

1 4


= 5(4) – 1(–3) = 23 = area

Note that if a, b, c and d are integers,

then the -area is an integer also.

Here is another motivation for the 2x2 determinants.

Determinant


Determinant

Il. (Algebraic) The system of equations

deta b

c d ≠ 0

ax + by = #

cx + dy = #

has exactly one solution if and only if


Here are examples where we don’t have exactly

one solution.

Determinant


deta b

c d ≠ 0

ax + by = #

cx + dy = #


{ x + y = 2

2x + 2y = 4



one solution.

Duplicated information.

Infinitely many solutions.

Determinant


deta b

c d ≠ 0

ax + by = #

cx + dy = #


{ x + y = 2

2x + 2y = 4



one solution.



Determinant


deta b

c d ≠ 0

ax + by = #

cx + dy = #


{ x + y = 2

2x + 2y = 4 { x + y = 2

2x + 2y = 5



one solution.



Determinant


deta b

c d ≠ 0

ax + by = #

cx + dy = #


{ x + y = 2

2x + 2y = 4 { x + y = 2

2x + 2y = 5

Contradictory information.

No solutions.



one solution.



Determinant


deta b

c d ≠ 0

ax + by = #

cx + dy = #


{ x + y = 2

2x + 2y = 4 { x + y = 2

2x + 2y = 5

Contradictory information.

No solutions.

Note that the coefficient matrix has det1 12 2 = 0.

Here is the Cramer’s rule that actually gives

the only solution of the system

Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

{ x + y = 2

x + 4y = 5

Example B. Use the Cramer’s rule to solve for x & y.



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

{ x + y = 2

x + 4y = 5

Example B. Use the Cramer’s rule to solve for x & y.1 1

1 4 = 3 = D ≠ 0det



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

namely that x =det

u b v d

D

{ x + y = 2

x + 4y = 5


1 4 = 3 = D ≠ 0 det



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

namely that x =det

u b v d

D

{ x + y = 2

x + 4y = 5


1 4 = 3 = D ≠ 0 det

so that x =det

2 15 4

3



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

namely that x =det

u b v d

D

{ x + y = 2

x + 4y = 5


1 4 = 3 = D ≠ 0 det

so that x =det

2 15 4

3= 1



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

namely that x =det

u b v d

Dy =

deta uc v

D

{ x + y = 2

x + 4y = 5


1 4 = 3 = D ≠ 0 det

so that x =det

2 15 4

3= 1



Determinant

if deta b

c d = D ≠ 0

ax + by = u

cx + dy = v

namely that x =det

u b v d

Dy =

deta uc v

D

{ x + y = 2

x + 4y = 5


1 4 = 3 = D ≠ 0 det

so that x =det

2 15 4

3y =

det1 21 5

3= 1 = 1

The determinant of a 3×3 matrix, may be defined

in the following expansion of 2×2 determinants:

Determinant

det

a b c

d e f

g h i



Example C. Calculate.

Determinant

det

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

dete f

h i= a*



Determinant

det

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1


dete f

h i= a*



Determinant

det

a b c

d e f

g h i

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1


dete f

h i= a*



Determinant

det

a b c

d e f

g h i

det– b*d f

g i

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1


dete f

h i= a*



Determinant

det

a b c

d e f

g h i

det– b*d f

g i

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1


det– 0*2 0

2 –1

dete f

h i= a*



Determinant

det

a b c

d e f

g h i

det– b* det+ c*d f

g i

d e

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1


det– 0*2 0

2 –1

dete f

h i= a*




Determinant

det

a b c

d e f

g h i


g i

d e

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1

det– 0* det+ 2*2 0

2 –1

2 1

dete f

h i= a*




Determinant

det

a b c

d e f

g h i


g i

d e

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1

det– 0* det+ 2*2 0

2 –1

2 1

= 1(–1 – 0) – 0 + 2(6 – 2)

dete f

h i= a*




Determinant

det

a b c

d e f

g h i


g i

d e

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1

det– 0* det+ 2*2 0

2 –1

2 1

= 1(–1 – 0) – 0 + 2(6 – 2)

= –1 + 8

= 7

Don’t remember this!

dete f

h i= a*




Determinant

det

a b c

d e f

g h i

det– b* det+ c*

= a(ei – hf) – b(ei – hf) + c(dh – ge)

d f

g i

d e

det1 0

3 –1= 1*det

1 0 2

2 1 0

2 3 –1

det– 0* det+ 2*2 0

2 –1

2 1

= 1(–1 – 0) – 0 + 2(6 – 2)

= –1 + 8

= 7

Here is the useful Butterfly method

for the determinant of a 3×3 matrix.

Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det

Example D. Calculate using the Butterfly method.



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

Enlarge the matrix

by adding the

first two columns.



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

1 0 2

2 1 0

2 3 –1

1 0

2 1

2 3

Enlarge the matrix

by adding the

first two columns.



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

1 0 2

2 1 0

2 3 –1

1 0

2 1

2 3

Enlarge the matrix

by adding the

first two columns.

Add all these diagonal products



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

1 0 2

2 1 0

2 3 –1

1 0

2 1

2 3

Enlarge the matrix

by adding the

first two columns.


Add –1 0 12



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

1 0 2

2 1 0

2 3 –1

1 0

2 1

2 3

Enlarge the matrix

by adding the

first two columns.

Subtract all these diagonal products


Add –1 0 12



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

1 0 2

2 1 0

2 3 –1

1 0

2 1

2 3

Enlarge the matrix

by adding the

first two columns.



Subtract 4 0 0

Add –1 0 12



Determinant

a b c

d e f

g h i

det

1 0 2

2 1 0

2 3 –1

det


a b c

d e f

g h i

a b

d e

g h

1 0 2

2 1 0

2 3 –1

1 0

2 1

2 3

Enlarge the matrix

by adding the

first two columns.



Subtract 4 0 0

Add –1 0 12

(–1 +12) – 4 = 7

The det A of a 3×3 matrix A similarly gives

the volume of a solid in 3D space.

Determinant

Determinant

(a, b, c)

(d, e, f)

(0, 0, 0)

(g, h, i)

a b c

d e f

g h idet = ± the volume of



Specifically,

Determinant

(a, b, c)

(d, e, f)

(0, 0, 0)

(g, h, i)a b c

d e f




Specifically,

Again there the ± are the two “orientations” of the

objects defined by these coordinates.

Determinant

(a, b, c)

(d, e, f)

(0, 0, 0)

(g, h, i)a b c

d e f




Specifically,



If we switch two of the coordinates in our labeling,

say x and y, we would get a left handed copy

of the solid.

Determinant

(a, b, c)

(d, e, f)

(0, 0, 0)

(g, h, i)a b c

d e f




Specifically,



If we switch two of the coordinates in our labeling,

say x and y, we would get a left handed copy

of the solid. But if we switch the labeling again,

we would get back to the original right handed solid.

Determinant

(a, b, c)

(d, e, f)

(0, 0, 0)

(g, h, i)a b c

d e f


Likewise the 3x3 system

Determinant

≠ 0det

a b c

d e f

g h ihas a unique answer iff the

ax + by + cz = #

dx + ey + fz = #

gx + hy + iz = #


Determinant

There is a 3x3 (in general nxn) Cramer’s Rule that

gives the solutions for x, y, and z but it’s not useful

for computational purposes-too many steps!

≠ 0det

a b c

d e f


ax + by + cz = #

dx + ey + fz = #

gx + hy + iz = #


Determinant

ax + by + cz = #




dx + ey + fz = #

gx + hy + iz = #

≠ 0det

a b c

d e f


We will show a method of finding the determinants of

n x n matrices.


Determinant

ax + by + cz = #




dx + ey + fz = #

gx + hy + iz = #

≠ 0det

a b c

d e f


We will show a method of finding the determinants of

n x n matrices. The Cramer’s Rule establishes that:

an nxn linear system has a unique answer iff

the det(the coefficient matrix) ≠ 0.

DeterminantWe need two points to define the

two edges of a parallelogram



and three points for the three edges

of a tilted box.




of a tilted box. The coordinates of

these points form 2x2 and 3x3

square matrices and the determinants

of these matrices give the signed

areas and volumes respectively.









Likewise, we need n points to

determine an “n-dimensional box”

whose coordinates form an nxn matrix.









Likewise, we need n points to

determine an “n-dimensional box”

whose coordinates form an nxn matrix.

Below we give one method to find the determinants of

nxn matrices which is the signed “volumes” of these

“n-dimensional” boxes.

DeterminantFinding Determinants-Expansion by the First Row


Let A be an nxn matrix as shown.

a11 a12 . . a1n

a21 a22 . . a2n

. . . aij .

. . . . .

an1 . . . ann

A =



a11 a12 . . a1n

a21 a22 . . a2n

. . . aij .

. . . . .

an1 . . . ann

The submatrix Aij of A

is the matrix left after deleting

the i'th row and j’th column of A.A =



a11 a12 . . a1n

a21 a22 . . a2n

. . . aij .

. . . . .

an1 . . . ann



the i'th row and j’th column of A.

a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann

A =

Aij = aij



a11 a12 . . a1n

a21 a22 . . a2n

. . . aij .

. . . . .

an1 . . . ann




For example, if

a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann

A =

Aij =

A =1 0 2

2 1 0

2 3 –1, then

A23 =aij



a11 a12 . . a1n

a21 a22 . . a2n

. . . aij .

. . . . .

an1 . . . ann




For example, if

a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann

A =

Aij =

A =1 0 2

2 1 0

2 3 –1, then

A23 =

1 0 2

2 1 0

2 3 –1

delete the 2nd row

and the 3rd column

aij



a11 a12 . . a1n

a21 a22 . . a2n

. . . aij .

. . . . .

an1 . . . ann




For example, if

a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann

A =

Aij =

A =1 0 2

2 1 0

2 3 –1, then

A23 =

1 0 2

2 1 0

2 3 –1

delete the 2nd row

and the 3rd column

aij=

1 0

2 3




a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann

A =



a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann

The determinant of A or det(A)

may be calculated using the

determinants of its

submatrices Aij’s. A =



a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann



determinants of its

submatrices Aij’s.

Here is the formula for det(A)

using the determinants of the

submatrices from the 1st row.

A =



a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann



determinants of its





A =

det(A) ≡ a11det(A11)–a12det(A12) +a13det(A13) – ..(±)a1ndet(A1n)



a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann



determinants of its





A =


Hence using this formula,

det1 0 2

2 1 0

2 3 –1



a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann



determinants of its





A =



det1 0 2

2 1 0

2 3 –1

= 1det – 0 + 2det 1 03 –1

2 12 3



a11 a12 . . a1n

a21 a22 . . a2n

. . . . .

. . . . .

an1 . . . ann



determinants of its





A =



det1 0 2

2 1 0

2 3 –1

= 1det – 0 + 2det 1 03 –1

2 12 3

= –1 + 8 = 7 (same answer as in eg. D)

DeterminantExample E. Find the following determinant using

the 1st row expansion.

2 0 −1 01 0 0 20 1 1 0−1 2 0 1

det



2 0 −1 01 0 0 20 1 1 0−1 2 0 1

det

= 2 det

0 0 2

1 1 0

2 0 1

2



2 0 −1 01 0 0 20 1 1 0−1 2 0 1

det

= 2 det

0 0 2

1 1 0

2 0 1

– 0

0



2 0 −1 01 0 0 20 1 1 0−1 2 0 1

det

= 2 det

0 0 2

1 1 0

2 0 1

– 0 + (−1)det1 0 2

0 1 0

–1 2 1

−1



2 0 −1 01 0 0 20 1 1 0−1 2 0 1

det

= 2 det

0 0 2

1 1 0

2 0 1

– 0 + (−1)det1 0 2

0 1 0

–1 2 1

0

– 0



2 0 −1 01 0 0 20 1 1 0−1 2 0 1

det

= 2 det

0 0 2

1 1 0

2 0 1

– 0 + (−1)det1 0 2

0 1 0

–1 2 1

– 0

= −11

Education

6.5 determinant x