Linear Algebra Exercises-n-Answers.pdf

118 Linear Algebra, by Hefferon
As a check, note that the third column of the starting matrix is 3/2 times the second, and so it is
indeed singular and therefore has no inverse.
Three.IV.4.17 We can use Corollary 4.12.
1
1 · 5 − 2 · 3 ·
( ) ( )
5 −3 −5 3
=
−2 1 2 −1
Three.IV.4.18 (a) The proof that the inverse is r −1 H −1 = (1/r) · H −1 (provided, of course, that
the matrix is invertible) is easy.
(b) No. For one thing, the fact that H + G has an inverse doesn’t imply that H has an inverse or
that G has an inverse. Neither of these matrices is invertible but their sum is.
( ) ( )
1 0 0 0
0 0 0 1
Another point is that just because H and G each has an inverse doesn’t mean H + G has an inverse;
here is an example.
( ) ( )
1 0 −1 0
0 1 0 −1
Still a third point is that, even if the two matrices have inverses, and the sum has an inverse, doesn’t
imply that the equation holds:
( ) −1 ( ) −1 ( ) −1 ( ) −1
2 0 1/2 0
3 0 1/3 0
=
=
0 2 0 1/2 0 3 0 1/3
but
and (1/2) + (1/3) does not equal 1/5.
( ) −1 ( ) −1
5 0 1/5 0
=
0 5 0 1/5
Three.IV.4.19 Yes: T k (T −1 ) k = (T T · · · T ) · (T −1 T −1 · · · T −1 ) = T k−1 (T T −1 )(T −1 ) k−1 = · · · = I.
Three.IV.4.20 Yes, the inverse of H −1 is H.
Three.IV.4.21 One way to check that the first is true is with the angle sum formulas from trigonometry.
(cos(θ1 ) ( )
+ θ 2 ) − sin(θ 1 + θ 2 ) cos θ1 cos θ
=
2 − sin θ 1 sin θ 2 − sin θ 1 cos θ 2 − cos θ 1 sin θ 2
sin(θ 1 + θ 2 ) cos(θ 1 + θ 2 ) sin θ 1 cos θ 2 + cos θ 1 sin θ 2 cos θ 1 cos θ 2 − sin θ 1 sin θ 2
( ) ( )
cos θ1 − sin θ
=
1 cos θ2 − sin θ 2
sin θ 1 cos θ 1 sin θ 2 cos θ 2
Checking the second equation in this way is similar.
Of course, the equations can be not just checked but also understood by recalling that t θ is the
map that rotates vectors about the origin through an angle of θ radians.
Three.IV.4.22
There are two cases. For the first case we assume that a is nonzero. Then
( ) ( )
−(c/a)ρ 1 +ρ 2 a b 1 0 a b 1 0
−→ =
0 −(bc/a) + d −c/a 1 0 (ad − bc)/a −c/a 1
shows that the matrix is invertible (in this a ≠ 0 case) if and only if ad − bc ≠ 0. To find the inverse,
we finish with the Jordan half of the reduction.
( )
( )
(1/a)ρ 1 1 b/a 1/a 0 −(b/a)ρ 2+ρ 1 1 0 d/(ad − bc) −b/(ad − bc)
−→
−→
(a/ad−bc)ρ 2
0 1 −c/(ad − bc) a/(ad − bc)
0 1 −c/(ad − bc) a/(ad − bc)
The other case is the a = 0 case. We swap to get c into the 1, 1 position.
( )
ρ 1↔ρ 2 c d 0 1
−→
0 b 1 0
This matrix is nonsingular if and only if both b and c are nonzero (which, under the case assumption
that a = 0, holds if and only if ad − bc ≠ 0). To find the inverse we do the Jordan half.
( )
( )
(1/c)ρ 1 1 d/c 0 1/c −(d/c)ρ 2+ρ 1 1 0 −d/bc 1/c
−→
−→
(1/b)ρ 2
0 1 1/b 0
0 1 1/b 0
(Note that this is what is required, since a = 0 gives that ad − bc = −bc).