Begin2.DVI

Chapter 10

I10-1. (c) AB=

[

36 59

11 18

]

(d) B−^1 A−^1 =

[

−18 59

11 − 36

]

I10-2.
AA−^1 =I
(AA−^1 )T=IT=I
(A−^1 )TAT=I multiply by(AT)−^1
(A−^1 )T=(AT)−^1

I10-4. (a)
If AB=BA, left multipley byA−^1
A−^1 AB=A−^1 BA
B=A−^1 BAright multiply byA−^1
BA−^1 =A−^1 BAA−^1
BA−^1 =A−^1 B

I10-5. If A is symmetric, then AT =A so that if (A−^1 )TAT = I one can write

(A−^1 )TA=I. Multiply this last equation on the right byA−^1 to obtain(A−^1 )T=A−^1

which showsA−^1 is symmetric.

I10-6. Left multiply both sides of equation byA−^1

I10-7. IfAB=AandBA=B, then one can write

AB=Aright multiply by A

(AB)A=A^2 associative property

A(BA) =A^2 propertiesBA=BandAB=A

AB=A^2

A=A^2

BA=B right multiply by B

(BA)B=B^2 associative property

B(AB) =B^2 propertiesAB=AandBA=B

BA=B^2

B=B^2

I10-8. (a) LetY =A^2 =AAwithY−^1 = (A^2 )−^1 , then one can write

I=Y Y−^1 = (A^2 )(A^2 )−^1 left multiply byA−^1

A−^1 =A−^1 AA(A^2 )−^1 left mulitply byA−^1

(A−^1 )^2 =A−^1 A(A^2 )−^1 = (A^2 )−^1

I10-9. (a) LetB=AAT, thenBT= (AAT)T=AAT=B

(d) A=^12 (A+AT) +^12 (A−AT)

Solutions Chapter 10