If $A'=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix}$ and $B=\begin{bmatrix}-1 & 0\\2 & 3\end{bmatrix}$
then $(3A+2B)'$ is :
Answer & explanation
Correct answer: option 3
$A'=\begin{pmatrix}-2&3\\1&2\end{pmatrix}$
$A=(A')^T=\begin{pmatrix}-2&1\\3&2\end{pmatrix}$
$B=\begin{pmatrix}-1&0\\2&3\end{pmatrix}$
$3A=\begin{pmatrix}-6&3\\9&6\end{pmatrix}$
$2B=\begin{pmatrix}-2&0\\4&6\end{pmatrix}$
$3A+2B=\begin{pmatrix}-8&3\\13&12\end{pmatrix}$
$(3A+2B)'=\begin{pmatrix}-8&13\\3&12\end{pmatrix}$
The value of $(3A+2B)'$ is $\begin{pmatrix}-8&13\\3&12\end{pmatrix}$.