Matrices and Determinants 3 Question 4
5. If $A=\begin{bmatrix}2 & -3 \\ -4 & 1\end{bmatrix}$, then adj $\left(3 A^{2}+12 A\right)$ is equal to
(a) $\begin{bmatrix}72 & -84 \\ -63 & 51\end{bmatrix}$
(b) $\begin{bmatrix}51 & 63 \\ 84 & 72\end{bmatrix}$
(c) $\begin{bmatrix}51 & 84 \\ 63 & 72\end{bmatrix}$
(d) $\begin{bmatrix}72 & -63 \\ -84 & 51\end{bmatrix}$
(2017 Main)
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Answer:
Correct Answer: 5. (b)
Solution:
- We have, $\quad A=\begin{bmatrix}2 & -3 \\ -4 & 1\end{bmatrix}$
$ \begin{aligned} & \therefore \quad A^{2}=A \cdot A=\begin{bmatrix} 2 & -3 & 2 & -3 \\ -4 & 1 & -4 & 1 \end{bmatrix} \\ & =\begin{bmatrix} 4+12 & -6-3 \\ -8-4 & 12+1 \end{bmatrix} \\ & =\begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix} \end{aligned} $
Now, $3 A^{2}+12 A=3 \begin{bmatrix}16 & -9 \\ -12 & 13\end{bmatrix}+12 \begin{bmatrix}2 & -3 \\ -4 & 1\end{bmatrix}$
$ \begin{aligned} & =\begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix}+\begin{bmatrix} 24 & -36 \\ -48 & 12 \\ 72 & -63 \end{bmatrix} \\ & =\begin{bmatrix} -84 & 51 \end{bmatrix} \end{aligned} $
$\therefore \quad \operatorname{adj}\left(3 A^{2}+12 A\right)=\begin{bmatrix}51 & 63 \\ 84 & 72\end{bmatrix}$
