Question
Given $A=\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]$ and $A^t$ is its transpose matrix. Find $\frac{1}{2} A-\frac{1}{3} A^t$
✓
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] $
$ \frac{1}{2} A-\frac{1}{3} A^t$
$\begin{array}{l}=\frac{1}{2}\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]-\frac{1}{3}\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}\frac{-3}{2} & 3 \\ 0 & \frac{-9}{2}\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 2 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}\frac{-1}{2} & 3 \\ -2 & \frac{-3}{2}\end{array}\right]$
$A^t=\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] $
$ \frac{1}{2} A-\frac{1}{3} A^t$
$\begin{array}{l}=\frac{1}{2}\left[\begin{array}{cc}-3 & 6 \\ 0 & -9\end{array}\right]-\frac{1}{3}\left[\begin{array}{cc}-3 & 0 \\ 6 & -9\end{array}\right] \end{array} $
$ =\left[\begin{array}{cc}\frac{-3}{2} & 3 \\ 0 & \frac{-9}{2}\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 2 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}\frac{-1}{2} & 3 \\ -2 & \frac{-3}{2}\end{array}\right]$
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