If $A=\begin{bmatrix}1&2\\3&4\\5&6\end{bmatrix}$ and $B=\begin{bmatrix}-3&-2\\1&-5\\4&3\end{bmatrix}$, then find $D=\begin{bmatrix}p&q\\r&s\\t&u\end{bmatrix}$ such that $A+B-D=O$.
Answer & explanation
Correct answer: option 3
Here $A+B-D=O$.
$∴D=A+B$
$⇒\begin{bmatrix}p&q\\r&s\\t&u\end{bmatrix}=\begin{bmatrix}1&2\\3&4\\5&6\end{bmatrix}+\begin{bmatrix}-3&-2\\1&-5\\4&3\end{bmatrix}$
$\begin{bmatrix}1-3&2-2\\3+1&4-5\\5+4&6+3\end{bmatrix}=\begin{bmatrix}-2&0\\4&-1\\9&9\end{bmatrix}$
$⇒D=\begin{bmatrix}-2&0\\4&-1\\9&9\end{bmatrix}$