If $A^{\prime}=\left[\begin{array}{cc}-2 & 3 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{rr}-1 & 0 \\ 1 & 2\end{array}\right]$, then find $(A+2 B)^{\prime}$
We know that $A=\left(A^{\prime}\right)^{\prime}$
$\therefore A=\left[\begin{array}{rr}-2 & 1 \\ 3 & 2\end{array}\right]$\
$\therefore A+2 B=\left[\begin{array}{rr}-2 & 1 \\ 3 & 2\end{array}\right]+2\left[\begin{array}{rr}-1 & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{rr}-2 & 1 \\ 3 & 2\end{array}\right]+\left[\begin{array}{rr}-2 & 0 \\ 2 & 4\end{array}\right]=\left[\begin{array}{rr}-4 & 1 \\ 5 & 6\end{array}\right]$
$\therefore(A+2 B)^{\prime}=\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right]$
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