A) \[\left[ \begin{matrix} 72 & -63 \\ -84 & 51 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 72 & -84 \\ -63 & 51 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 51 & 63 \\ 84 & 72 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 51 & 84 \\ 63 & 72 \\ \end{matrix} \right]\]
Correct Answer: C
Solution :
Given \[A=\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \\ \end{matrix} \right]\] \[3{{A}^{2}}=\left[ \begin{matrix} 16 & -9 \\ -12 & 13 \\ \end{matrix} \right]\] \[12A=\left[ \begin{matrix} 24 & -36 \\ -48 & 12 \\ \end{matrix} \right]\] \[\therefore \] \[3{{A}^{2}}+12A=\left[ \begin{matrix} 72 & -63 \\ -84 & 51 \\ \end{matrix} \right]\] adj\[(3{{A}^{2}}+12A)=\left[ \begin{matrix} 51 & 63 \\ 84 & 72 \\ \end{matrix} \right]\]
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