Question

If $A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$, then $\left[3A^2+12A\right]$ is equal to

• A. $\left[\begin{array}{cc}72& -84\\ -63& 51\end{array}\right]$
• B. $\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]$
• C. $\left[\begin{array}{cc}51& 84\\ 63& 72\end{array}\right]$
• D. $\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}72& -84\\ -63& 51\end{array}\right]$

We have,

$A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$

$A^2=A\times A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]\times \left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]$

Thus,

$3A^2=\left[\begin{array}{cc}48& -27\\ -36& 39\end{array}\right]$

Now, $12A=\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]$

Therefore,

$[3A^2+12A]=\left[\begin{array}{cc}48& -27\\ -36& 39\end{array}\right]+\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right][3A^2+12A]=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$