Question
Given $A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ is its transpose matrix Find $A^t-\frac{1}{3} A$

Answer

$A^t-\frac{1}{3} A  $
$ =\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-\frac{1}{3}\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]  $
$=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-\left[\begin{array}{cc}-1 & 2 \\ 0 & -3\end{array}\right]  $
$ =\left[\begin{array}{cc}-2 & -2 \\ 6 & -6\end{array}\right]$

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