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Jeyner Cespedes Section 2.2 Pg 109 – 110 ex 27-‐37 odd Section 2.3 Pg 115 ex 1-‐9 odd 27. Let A be a 3 x 3 matrix. a. Use equation (2) from Section 2.1 to show that 𝒓𝒐𝒘𝒊 (A) = 𝒓𝒐𝒘𝒊 (I). A, for
i = 1,2,3.
Solution: 𝑟𝑜𝑤! (AB) = 𝑟𝑜𝑤! (A) x B Interchange A and B 𝑟𝑜𝑤!(BA) = 𝑟𝑜𝑤!(B) x A B gets replace by the identity matrix (I) 𝑟𝑜𝑤!(IA)= 𝑟𝑜𝑤!(I) x A 𝑟𝑜𝑤!(A) = 𝑟𝑜𝑤! (I) x A (we know that IA=A) Therefore 𝑟𝑜𝑤! 𝐴 = 𝑟𝑜𝑤! (I) x A 𝑟𝑜𝑤!(A)= 𝑟𝑜𝑤!(I)x A for i =1,2,3
𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)
= 𝑟𝑜𝑤! I x A𝑟𝑜𝑤! I 𝑥 𝐴𝑟𝑜𝑤! I x A
b. Show that if row 1 and 2 of A are interchanged, then the result may be written as EA, where E is an elementary matrix formed by interchanging rows 1 and 2 of I.
Solution:
𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)
= 𝑟𝑜𝑤! I x A𝑟𝑜𝑤! I 𝑥 𝐴𝑟𝑜𝑤! I x A
Row 1 and 2 of A are interchanged
𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)
= 𝑟𝑜𝑤! I x A𝑟𝑜𝑤! I 𝑥 𝐴𝑟𝑜𝑤! I x A
= 𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)
x A = EA
We get E by interchanging row 1 and 2. 𝑟𝑜𝑤!(EA)= 𝑟𝑜𝑤!(E) x A
c. Show that if row 3 of A is multiplied by 5, then the result may be written as EA, where E is formed by multiplying row 3 of I by 5.
Solution:
𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)
= 𝑟𝑜𝑤! I x A𝑟𝑜𝑤! I 𝑥 𝐴𝑟𝑜𝑤! I x A
row 3 of A is multiplied by 5
𝑟𝑜𝑤! (A)𝑟𝑜𝑤! (A)
5 𝑥 𝑟𝑜𝑤! (A)=
𝑟𝑜𝑤! I x A𝑟𝑜𝑤! I 𝑥 𝐴
5 𝑥 𝑟𝑜𝑤! I x A=
𝑟𝑜𝑤! (I)𝑟𝑜𝑤! (I)
5 𝑥 𝑟𝑜𝑤! (I)A = EA
Find the inverses of the matrices in Exercises 29-‐32, if they exist. Use the algorithm introduced in this section. 29. 1 −3
4 −9 = 1 −3 14 −9 0
01
−4 𝑅1+R2=R2 1 −3 10 3 −4
01
(1/3)X R2 = R2 1 −3 10 1 −4/3
01/3
3 X R2 + R1 = R1 1 0 −30 1 −4/3
11/3
Is it invertible 𝐴!! = −3 1−4/3 1/3
31. 1 0 −2−3 1 42 −3 4
A I = 1 0 −2−3 1 42 −3 4
100 0 01 00 1
3R1+R2= R2 (-‐2)R1+R3= R3 1 0 −20 1 −20 −3 8
13−2 0 01 00 1
3R2+R3=R3 1 0 −20 1 −20 0 2
137 0 01 03 1
R3+R1= R1 R3+R2=R2 1 0 00 1 00 0 2
8107 3 14 13 1
(½) R3=R3 1 0 00 1 00 0 2
8107/2
3 14 13/2 1/2
It is invertible 𝐴!!= 8107/2
3 14 13/2 1/2
33. Use the algorithm from this section to find the inverses of
I.
A= 1 0 01 1 01 1 1
x (I)
1 0 01 1 01 1 1
1 0 00 1 00 0 1
R2-‐R1=R2 R3-‐R1=R3 1 0 00 1 00 1 1
1 0 0−1 1 0−1 0 1
R3-‐R2=R3 1 0 00 1 00 0 1
1 0 0−1 1 00 −1 1
It is invertible 𝐴!! = 1 0 0−1 1 00 −1 1
II.
B=
1 0 0 01 1 0 01 1 1 01 1 1 1
x (I)
1 0 0 0 1 0 0 01 1 0 0 0 1 0 01 1 1 0 0 0 1 01 1 1 1 0 0 0 1
R2-‐R1 R3-‐R1 R4-‐R1 1 0 0 0 1 0 0 00 1 0 0 −1 1 0 00 1 1 0 −1 0 1 00 1 1 1 −1 0 0 1
R3-‐R2 R4-‐R2 1 0 0 0 1 0 0 00 1 0 0 −1 1 0 00 0 1 0 0 −1 1 00 0 1 1 0 −1 0 1
R4-‐R3 1 0 0 0 1 0 0 00 1 0 0 −1 1 0 00 0 1 0 0 −1 1 00 0 0 1 0 0 −1 1
It is invertible 𝐵!! = 0 0 0 0−1 1 0 00 −1 1 00 0 −1 1
The inverse of nxn matrix
B=
0 0 0 0 0−1 1 0 0 00 −1 1 0 0. . . . .. . . . .. . . . .0 0 0 1 1
35. Let A= −1 −7 −32 15 61 3 2
. Find the third column of 𝐴!! without computing the
other columns. −1 −7 −3 1 0 02 15 6 0 1 01 3 2 0 0 1
2R1+R2=R2 −1 −7 −3 1 0 00 1 0 2 1 01 3 2 0 0 1
1R1+R3=R3 −1 −7 −3 1 0 00 1 0 2 1 00 −4 −1 1 0 1
𝐴!!= −30−1
37. Let A = 1 21 31 5
. Construct a 2 x 3 matrix C (by trial and error) using only 1, -‐1,
and 0 as entries, such that CA= 𝐼!. Compute AC and note that AC ≠ 𝐼!.
𝐶𝐴 =𝑎! 𝑎! 𝑎!𝑏! 𝑏! 𝑏!
1 21 31 5
𝑎! + 𝑎! + 𝑎! 2𝑎! + 3𝑎! + 5𝑎!𝑏! + 𝑏! + 𝑐! 2𝑏! + 3𝑏! + 5𝑏!
= 1 00 1
𝑎! + 𝑎! + 𝑎! 0
0 2𝑏! + 3𝑏! + 5𝑏!
C= 1 1 −1
−1 1 0
Useless otherwise specified, assume that all matrices in these exercises are n x n. Determine which of the matrices in Exercises 1-‐10 are invertible. Use as few calculations as possible. 1. 5 7
−3 −6 5 7 1 0−3 −6 0 1
(1/5)R1=R1
175
15 0
−3 −6 0 1
3R1+R2=R2
175
15 0
0 3 3/5 1
It is invertible because it has pivot point in every row
3. 3 0 0−3 −4 08 5 −3
𝐴! has pivot in every column so it is invertible
5. 3 0 −32 0 4−4 0 7
Not invertible since this matrix has a column of zeros
7.
−13−20
−3 0 15 8 −3−6 3 2−1 2 1
3R1+R2=R2 R3-‐2R1=R3 −1000
−3 0 1−4 8 00 3 01 2 1
R4 -‐ (1/4)R2 = R4 −1000
−3 0 1−4 8 00 3 00 0 1
9. [M] 4−67−1
0 −3 −79 9 9−5 10 192 4 −1
It has four positions, so it is invertible
