2

So this is a homework problem. So I'll give what I got.

Suppose $A,$ $\ B$ & $X$ are $n$ by $n$ matrices with $A,$ $\ X$ & $A-AX$ invertible. Now suppose that $$(1) \quad (A-AX)^{-1}=X^{-1}B$$

$A)$ Explain why $B$ is invertible.

$B)$ Solve for $X$.

Part A:

Multiply both sides of $(1)$ by $B^{-1}$ to get,

$$(2) \quad (A-AX)^{-1}B^{-1}=X^{-1}$$ Now Multiply $(2)$ by $(A-AX)^{-1}$ to get,

$$(3) \quad B^{-1}=(A-AX)X^{-1}$$

$(A-AX)$ exists, and the existence of $X^{-1}$ is a given. The matrices are of the same size so matrix multiplication is defined. Therefore $B^{-1}$ exists.

Part B

From $(3)$ use the fact that,

$C^{-1}D^{-1}=(DC)^{-1}$

To manipulate the RHS of $(3)$ into,

$$(4) \quad B^{-1}=(X(A-AX)^{-1})^{-1}$$ $$\Rightarrow B^{-1}=(X(A(I-X))^{-1})^{-1}$$ $$\Rightarrow B^{-1}=(X(I-X)^{-1}(A)^{-1})^{-1}$$ $$\Rightarrow B^{-1}=(((I-X)X^{-1})^{-1}A^{-1})^{-1}$$ $$\Rightarrow B^{-1}=A(I-X)X^{-1}$$ $$\Rightarrow B^{-1}=A(X^{-1}-I)$$ $$\Rightarrow A^{-1}B^{-1}=(X^{-1}-I)$$ $$\Rightarrow X=(BA)^{-1}+I$$

Seems correct to me. But I'm a bit bias. Is this correct? If not, where's the error?

Zach466920
  • 8,341

3 Answers3

3

Your part A is problematic. You can't start with $B^{-1}$ until you have proved that $B$ is invertible.

Approach 1: propose a candidate and show that it works. Let $M=(A-AX)X^{-1}$. Then $$ MB=(A-AX)X^{-1}B=(A-AX)(X^{-1}B)=(A-AX)(A-AX)^{-1}=I, $$ and then use this result to infer that $BM=I$ also.

Approach 2: Take determinant on both sides of $(A-AX)^{-1}=X^{-1}B$ to infer that $\det(B)=\det X/\det(A-AX)\neq 0$.

Community
  • 1
yurnero
  • 10,505
2

Your answer to Part A is not right, because by multiplying by $B^{-1}$, you assumed it exists. Instead, starting from $$ (A-AX)^{-1}=X^{-1}B, \tag{1}$$ multiply by $X$ on the left, which gives $$ B= X(A-AX)^{-1}. $$ Then $B$ is a product of invertible matrices (according to the question) and so $B$ is invertible.

A shorter way to do Part B is to multiply the above formula on the right by $A-AX$. Then $$ BA(I-X)=X, $$ or $$ BA=(I+BA)X, $$ and then $$ X=(I+BA)^{-1}BA $$ (there are probably other ways of writing this, though).

(And by the way, when multiplying by a non-commutative element, you should state on which side you are placing it, since otherwise it can be ambiguous.)

Chappers
  • 67,606
0

In part a) you cannot multiply by $B^{-1}$ because you don't know whether it is invertible or not. You need to write the solution for part a) as: $$(A-AX)^{-1}=X^{-1}B$$ implies that $$X(A-AX)^{-1}=B=(X^{-1})^{-1}(A-AX)^{-1}=((A-AX)X^{-1})^{-1}$$ so the matrix $B$ is the inverse of matrix $(A-AX)X^{-1}$. $$$$ Hence B is invertible. $$$$ The part B seems to be right!

Kerr
  • 1,866