Problem 2. Use a combination of column and row operations to calculate the inverse A−1 of


1 1
x 1
x 1
x
A= 0
x2
Answer

h
A
x
i 
I3 = 
0
x2

x

r1 ↔ r3 0
x2
1
1 1
0
x
1
0
1
x
1
0
0
1 1 1
x 1 0
x 1 0
 2
x
1
r3 − r1 → r3  0
x
0
 2
x
x

r3 → r3 0
x−1
0
x
x
0
1
1
x−1
x
x 1
x 1
0 1

0

0

1
0
1
0

0
0
1
0
0
1
0
1
0

1
0 
− x1

0
1
1
0 
1
0 − x−1
0
0
x
x−1
 2
x
r2 − r3 → r2  0
0
x 1
0
x
x 0 − x−1
x
0 1
x−1
0
1
0
1 
x−1
1
− x−1
 2
x
r1 − r3 → r1  0
0
x
x 0 − x−1
x
x 0 − x−1
x
0 1
x−1
0
1
0
x 
x−1
1 
x−1
1
− x−1
x
x 0 − x−1
1
1 0 − x−1
x
0 1
x−1
 2
x
1

r2 → r2 0
x
0
 2
x
r1 − x r2 → r1  0
0
0 0
1 0
0 1

1 0 0
1
0 1 0
r
→
r
1
1
x2
0 0 1

1
x
x
x−1
0
− x12
1
1
x
x
x−1
0
1
0
1
− x−1
x
x−1
A−1 = − x−1
1
x
−1
1
− x−1
0

x
x−1
1

x(x−1)
1
− x−1
0
0
0
− x12
1
x
0

1
1

1

x(x−1)
1
− x−1

1
x2
1

x(x−1)
1
− x−1

1
x2
1

x(x−1)
1
− x−1
Problem 3. Calculate the L, U -factorization of

1 0
A= 0 1
2 2

−1
3
2


−1
3
2
Answer
1 0
A= 0 1
2 2

1
r2 − 2 r1 → r2 ∼ 0
0
0
1
2

−1
3
4
then L31 = 2

1
r3 − 2 r2 → r3 ∼ 0
0
0
1
0

−1
3 =U
−2
then L32 = 2

1
L = 0
2

0
0
1
0
1
2

1
−1
3  0
2
−2

1 0
A = LU = 0 1
0 0
0
1
2

0
0
1
Check

1
0
2
0
1
2

1
0
0 0
0
1
0
1
0

−1
3 =
−2


1 · 1 + 0 · 0 + 0 · 0 1 · 0 + 0 · 1 + 0 · 0 1 · (−1) + 0 · 3 + 0 · (−2)
0 · 1 + 1 · 0 + 0 · 0 0 · 0 + 1 · 1 + 0 · 0 0 · (−1) + 1 · 3 + 0 · (−2) =
2 · 1 + 2 · 0 + 1 · 0 2 · 0 + 2 · 1 + 1 · 0 2 (−1) + 2 · 3 + 1 · (−2)

1
0
2
0
1
2

−1
3 =A
2
2
Problem 4. A n × n permutation matrix P is any matrix obtained by interchanging a finite amount of rows (or
columns) of the n × n identity matrix I. Verify that P 2 = I in the 3 × 3 case by finding all six 3 × 3 permutation
matrices (hint, one of them is I) and calculate each of their inverses
Answer

1
P1 = 0
0

1
(P1 )2 = 0
0
0
0
1

0 1
1 0
0 0
0
0
1
 
0
1· 1+0· 0+0· 0
1 = 0 · 1 + 0 · 0 + 1 · 0
0
0· 1+1· 0+0· 0

0
1
0
0
0
1
1· 0+0· 0+0· 1
0· 0+0· 0+1· 1
0· 0+1· 0+0· 1
1· 0+0· 1+0·
0· 0+0· 1+1·
0· 0+1· 1+0·
 
0
1
0 = 0
0
0
0
1
0

0
0 = I
1
then P1−1 = P1

0
P2 = 1
0

0
(P2 )T = 0
1

0
(P2 )(P2 )T = 1
0
0
0
1

1 0
0 0
0 1
1
0
0

1
0
0
0
0
1
1
0
0
 
0
0· 0+0· 0+1· 1
1 = 1 · 0 + 0 · 0 + 0 · 1
0
0· 0+1· 0+0· 1

0
1
0
0· 1+0· 0+1· 0
1· 1+0· 0+0· 0
0· 1+1· 0+0· 0
 
0· 0+0· 1+1· 0
1
1 · 0 + 0 · 1 + 0 · 0 = 0
0· 0+1· 1+0· 0
0
0
1
0

0
0 = I
1
then (P2 )−1 = (P2 )T

0
P3 = 1
0

0
(P3 )2 = 1
0
1
0
0

0 0
0 1
1 0
1
0
0
 
0
0· 0+1· 1+0· 0
0 = 1 · 0 + 0 · 1 + 0 · 0
0· 0+0· 1+1· 0
1
1
0
0

0
0
1
0· 1+1· 0+0· 0
1· 1+0· 0+0· 0
0· 1+0· 0+1· 0
then (P3 )−1 = P3
3
 
0· 0+1· 0+0· 1
1
1 · 0 + 0 · 0 + 0 · 1 = 0
0· 0+0· 0+1· 1
0
0
1
0

0
0 = I
1

0
P4 = 0
1

0
(P4 )T = 1
0

0
(P4 )(P4 )T = 0
1
1
0
0

0
0
1 1
0
0
0
0
1

0
1
0
1
0
0
0
0
1

1
0
0
 
1
0· 0+1· 1+0· 0 0· 0+1· 0+0· 1 0· 1+1· 0+0·
0 = 0 · 0 + 0 · 1 + 1 · 0 0 · 0 + 0 · 0 + 1 · 1 0 · 1 + 0 · 0 + 1 ·
0
1· 0+0· 1+0· 0 1· 0+0· 0+0· 1 1· 1+0· 0+0·
 
0
1
0 = 0
0
0
0
1
0

0
0 = I
1
(P4 )−1 = (P4 )T


0 0 1
P5 =  0 1 0 
1 0 0

0
(P5 )2 = 0
1
0
1
0

1 0
0 0
0 1
0
1
0
 
1
0· 0+0· 0+1· 1
0 = 0 · 0 + 1 · 0 + 0 · 1
0
1· 0+0· 0+0· 1
0· 0+0· 1+1· 0
0· 0+1· 1+0· 0
1· 0+0· 1+0· 0
 
0
1
0 = 0
0
0
0
1
0

0
0 = I
1
 
1· 0+0· 0+0· 1
1
0 · 0 + 1 · 0 + 0 · 1 = 0
0· 0+0· 0+1· 1
0
0
1
0

0
0 = I
1
0· 1+0· 0+1·
0· 1+1· 0+0·
1· 1+0· 0+0·
(P5 )−1 = P5

1 0 0
P6 =  0 1 0 
0 0 1


1
(P6 )2 = 0
0
0
1
0

0 1
0 0
1 0
0
1
0
 
0
1· 1+0· 0+0· 0
0 = 0 · 1 + 1 · 0 + 0 · 0
1
0· 1+0· 0+1· 0
1· 0+0· 1+0· 0
0· 0+1· 1+0· 0
0· 0+0· 1+1· 0
(P6 )−1 = P6
4
Problem 6. Consider the following system of equations
x+3y − z = 1
y+z =2
4x+2y + 2z = 3
Solve this system of equations by calculating the inverse A−1 of the coefficient matrix A. What is the solution if
perturb the right hand side–i.e., what is the solution to
x+3y − z = 1 + 1
y + z = 2 + 2
4x+2y + 2z = 3 + +3
5