Question
Given $A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ is its transpose matrix Find $2 A^t-3 A$
✓
Answer
$\begin{array}{l}A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] \end{array} $
$ A^t=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]$
$2 A^t-3 A $
$ =2\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-3\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] $
$ =\left[\begin{array}{cc}-6 & 0 \\ 12 & -18\end{array}\right]-\left[\begin{array}{cc}-9 & 18 \\ 0 & -27\end{array}\right] $
$ =\left[\begin{array}{cc}3 & -18 \\ 12 & 9\end{array}\right]$
$ A^t=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]$
$2 A^t-3 A $
$ =2\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right]-3\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right] $
$ =\left[\begin{array}{cc}-6 & 0 \\ 12 & -18\end{array}\right]-\left[\begin{array}{cc}-9 & 18 \\ 0 & -27\end{array}\right] $
$ =\left[\begin{array}{cc}3 & -18 \\ 12 & 9\end{array}\right]$
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