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solve the following matrix equation for $X$:

$$ A(X-B)^{-1}=B$$ where

$$A = \left [\begin{array}{ccc} 1 & 2 \\ 3 & 4 \\ \end{array} \right ]$$ $$B = \left [\begin{array}{ccc} 1 & 1 \\ 1 & 3 \\ \end{array} \right ]$$

How I solved it:

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

Is that correct?

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  • $\begingroup$ What do you mean?? $\endgroup$ – hussi Mar 9 '15 at 20:23
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You have the right idea, but are off by a little bit. First, it is important to point out that $B$ is invertible because it is has a non-zero determinant. Then from the line $$A=B(X-B)=BX-B^2$$ you have to add $B^2$ to both sides to get $$A+B^2=BX$$ From here, you can multiply $B^{-1}$ (since we know it exists) to the left hand side, not the right hand side to get a final answer of $$B^{-1}(A+B^2)=X$$ It does matter which side you multiply on, because we don't know that $B$ or $B^{-1}$ can commute with $X$.

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  • $\begingroup$ Surely we have $A(X-B)^{-1}=B \implies A=B(X-B) \implies A = BX - B^2$? $\endgroup$ – Unochiii Mar 9 '15 at 20:33
  • $\begingroup$ Your a hero, thank you very much <3 $\endgroup$ – hussi Mar 9 '15 at 20:33
  • $\begingroup$ @Unochiii surely you are correct ;) Thanks for pointing that out. $\endgroup$ – graydad Mar 9 '15 at 20:34
  • $\begingroup$ @hussi no problem, but Unochiii pointed out one place I did make a mistake. The answer should be correct now that I've edited it. $\endgroup$ – graydad Mar 9 '15 at 20:35
  • $\begingroup$ so its not B^−1(A+B)=X? uh nvm i see it now :) $\endgroup$ – hussi Mar 9 '15 at 20:35

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