HW Solutions, 2.2
2.2, 16 Suppose that A, B are n × n, B and AB are invertible. Show that A is invertible.
SOLUTION: Using the hint, if C = AB, then C is invertible, and
CB −1 = ABB −1 = A
Therefore, A is the product of the invertible matrix C and B −1 , so A is invertible.
2.2, 18 Suppose P is invertible and A = P BP −1 . Solve for B in terms of A.
SOLUTION: Be sure you multiply on the correct side:
A = P BP −1 ⇒ P −1 A = P −1 P BP −1 = BP −1 ⇒ P −1 AP = BP −1 P = B
2.2, 20 Suppose that A, B, X are n × n with A, X, and A − AX invertible. Suppose that
(A − AX)−1 = X −1 B
(a) Show that B is invertible:
SOLUTION: Multiply on the left by X, and X(A − AX)−1 = B. Therefore, B is
the product of invertible matrices (and is itself invertible).
(b) Solve the equation given above for X. If you need to invert a matrix, explain why.
SOLUTION: I don’t recall what we did in class (there are multiple ways of expressing the solution), but here is one possibility. Multiply both sides by (A − AX):
X(A − AX)−1 = XX −1 B
X = B(A − AX) = BA − BAX
⇒
⇒
X(A − AX)−1 = B
X + BAX = BA
⇒
⇒
(I + BA)X = BA
Is I + BA invertible? The last equality implies that I + BA = BAX −1 , which is
the product of three invertible matrices. Therefore
X = BA(I + BA)−1
Alternative Solution: Another solution might be the following- First, invert both
sides of the given equation, since we know that B is invertible:
A − AX = B −1 X
A = B −1 X + AX = (B −1 + A)X
⇒
This last equation tells us that B −1 + A = AX −1 , so it is also invertible. Now do
the inversion:
X = (B −1 + A)−1 A
Just for fun: Can you make the two solutions look alike?
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