If $A=\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right]$, then adj $\left(3 A^2+12 A\right)$ is equal…
If $A=\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right]$, then adj $\left(3 A^2+12 A\right)$ is equal to
- $\left[\begin{array}{cc}-21 & 63 \\ 84 & 0\end{array}\right]$
- $\left[\begin{array}{cc}21 & 63 \\ 84 & 0\end{array}\right]$
- $\left[\begin{array}{cc}21 & -63 \\ 84 & 0\end{array}\right]$
- $\left[\begin{array}{cc}-21 & -63 \\ 84 & 0\end{array}\right]$
Solution
$\begin{aligned} & 3 A^2+12 A=3\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right]^2+12 \\ & =3\left[\begin{array}{cc}-8 & 9 \\ -12 & -11\end{array}\right]+12\left[\begin{array}{cc}2 & 3 \\ -4 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}-24+24 & 27+36 \\ -36-48 & -33+12\end{array}\right] \\ & =\left[\begin{array}{cc}0 & 63 \\ -84 & -21\end{array}\right] \\ & \Rightarrow \operatorname{adj}\left(3 A^2+12 A\right)=\left[\begin{array}{cc}-21 & -63 \\ 84 & 0\end{array}\right]\end{aligned}$
Asked in: MHT CET 2022 (08 Aug Shift 2)
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