I10-28. A−^1 =
1 0 0
−3 1 0
11 −5 1
B−^1 =
1 2 − 7 − 8
0 1 − 3 − 3
0 0 1 1
0 0 0 1
C−^1 =
8 −6 4 − 2
−6 12 − 8 − 4
4 −8 12 − 6
−2 4 −6 8
I10-29. IfAAT=
α^22 + 1/ 4 α^1 + 4 α^20
α 1 +α 2
2 α(^21) + 1/4 0
0 0 α^23
=
1 0 0
0 1 0
0 0 1
then one must selectα^21 =
α^22 = 3/ 4 ,α 1 =−α 2 andα^23 = 1
I10-30. (b)
d|A|
dt =−^4 t
(^3) + 6t (^2) − 6 t
I10-32. |A|= 1, |B|= 1, |A+B|= 3, |A|+|B|= 2
I10-36. (a)
eAt=I+At+A^2
t^2
∫ 2!+···
t
0
eAtdt=It+A
t^2
2!+A
2 t^3
3!+···
A
∫t
0
eAtdt=At+A^2
t^2
2!+A
3 t^3
3!+···
A
∫t
0
eAtdt+I=eAt
(b)
∫t
0
eAtdt=It+At
2
2!
+A^2 t
3
3!
+···
∫t
0
eAtdt=A−^1
[
At+A^2
t^2
2!+A
3 t^3
3!+···
]
= [At+A^2
t^2
2!+A
3 t^3
3!+···]A
− 1
∫t
0
eAtdt=A−^1
[
eAt−I
]
= [eAt−I]A−^1
I10-42. (b) yn+2− 6 yn+1+ 9yn= 0 Assume a solution yn=An with yn+1 =An+1
andyn+2=An+2, then substitute the assumed solution into the difference equation
and obtain the characteristic equation(A−3)^2 = 0with repeated rootsA= 3, 3. The
fundamental set of solutions is{ 3 n,n 3 n}and the general solution isyn=c 0 (3)n+c 1 n(3)n
Solutions Chapter 10