Question

Matrix $A=\left[a_{ij}\right],$ where $a_{ij}=(i+j{)}^2$. If $B=-3A$, find $B$.

• A. $\left[\begin{array}{cc}-12& -3\\ -3& -32\end{array}\right]$
• B. $\left[\begin{array}{cc}4& 9\\ 9& 16\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 9\\ 9& -16\end{array}\right]$
• D. $\left[\begin{array}{cc}-12& -27\\ -27& -48\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}-12& -27\\ -27& -48\end{array}\right]$

$A=\left[a_{ij}\right]$

Since, $a_{ij}={(i+j)}^2$

$\therefore$ $a_{11}={(1+1)}^2={\left(2\right)}^2=4,$

$a_{12}={(1+2)}^2={\left(3\right)}^2=9,$

$a_{21}={(2+1)}^2={\left(3\right)}^2=9,$

$a_{22}={(2+2)}^2={\left(4\right)}^2=16,$

So, matrix $A$ is $\left[\begin{array}{cc}4& 9\\ 9& 16\end{array}\right]$

Now, $B=-3A$ $=-3\left[\begin{array}{cc}4& 9\\ 9& 16\end{array}\right]$ $=\left[\begin{array}{cc}-12& -27\\ -27& -48\end{array}\right]$

$\therefore$ Option D is correct.