Solving linear matrix equation $BX=A-BX$

Question

Let $A,B,X$ ∈ $M_n$(ℝ), with $A, B$ invertible. In the following cases, find $X$ in terms of $A, B$ and their inverses.

(i) $BX=A-BX$

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For (i) I did

$BX=A-BX$ $\rightarrow$ $2BX=A$ $\rightarrow$ $2X=B^{-1}A$ $\rightarrow$ X = 1/2$B^{-1}A$=$X$.

Is this correct?

For (ii) $A-I_n$ is invertible.

(ii) $AX=X+B$

$AX=X+B$ $\rightarrow$ $AX-X=B$ $\rightarrow$ $(A-I_n)$$^{-1}$$X$=$B$ $\rightarrow$ $(A-I_n)$$^{-1}$$B$=$X$.

Is this correct?

(iii) $AXB$$^{-1}$ = A$^{-1}$+ $B$.

I am unsure what to do for (ii). Any help is greatly appreciated!

Answer 1

i)

$$BX=A-BX\implies 2BX=A\implies X=\frac{1}{2}B^{-1}A$$

ii)

$$AX=X+B\implies AX-X=B\implies (A-I_n)X=B\implies X=(A-I_n)^{-1}B$$

iii)$$AXB^{-1}=A^{-1}+B\implies AX=A^{-1}B+B^2\implies \\X=A^{-2}B+A^{-1}B^2=A^{-1}B(A^{-1}+B)$$

In i) i don't know what exactly you did since it does not make sense, you matrix $A$ simply vanishes and you end up with $X=(B-B)^{-1}$ which is zero despite $B$ goes from RHL to LHL adding

Answer 2

(i) it is not correct: from $BX=A-BX$ we get $2BX=A$ , hence $2X=B^{-1}A, $ thus

$$X= \frac{1}{2}B^{-1}A.$$

(ii) Since $AXB^{-1}=A^{-1}+B$ , it follows that $AX=A^{-1}B+B^2,$ thus

$$X=A^{-2}B+A^{-1}B^2.$$