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I don't have the answers to this but can someone correct I can't see the mistake.

Exercise

Andrei
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Maria
  • 15
  • The transpose of a matrix $A$ is denoted by $A^t$ or $A^T$, not $A^+$. The notation $A^+$ usually means something else, such as Moore-Penrose pseudoinverse. – user1551 Sep 16 '20 at 16:07
  • Check the 5th line, we have $A$ is symmetric so $A^T = A$ and $B$ is antisymmetric so $B^T = -B$. Also, check the last line. – MathLearner Sep 16 '20 at 21:57

3 Answers3

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The issue is on the last line. You need to multiply with $(A-I)^{-1}$ on the right. So you will get answer A

Andrei
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0

According to the law of transpose, from $(AX^T+B)^T=X+B)$ we get $XA^T+B^T=X+B$. Now $B^T=-B$ and $A^T=A$ so this reduces to $AX-B=X+B$ and $(A-I)X=2B$.

IF!!!!!!!! $A-I$ is regular, then $X=2(A-I)^{-1}B$ but it might not be the case (see $A=I$ for example).

YJT
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We have

\begin{equation} \begin{split} & (AX^T + B)^T & = X + B \\ \implies & XA^T + B^T & = X + B \\ \implies & XA^T - X & = -B^T + B \end{split} \end{equation}

We have $A$ is symmetric and $B$ is antisymmetric, so:

$$ A^T = A, \qquad \text{and} \qquad B^T = -B$$

Then

\begin{equation} \begin{split} & \qquad XA^T - X & = -B^T + B \\ & \implies XA - X & = B + B \\ & \implies X(A - I) & = 2B \\ & \implies X &= 2B(A - I)^{-1} \end{split} \end{equation}

MathLearner
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