Aspire Faculty ID #14053 · Topic: JAMIA MILLIA ISLAMIA MCA 2019 · Just now
JAMIA MILLIA ISLAMIA MCA 2019

If $A = \begin{bmatrix} 3 & -9 \\ -12 & 6 \end{bmatrix}$, then $\operatorname{adj}(3A + 12A^2)$ is equal to:

Solution

$A^2 = \begin{bmatrix} 3 & -9 \\ -12 & 6 \end{bmatrix}^2 = \begin{bmatrix} 3^2 + (-9)(-12) & 3(-9) + (-9)(6) \\ (-12)(3) + 6(-12) & (-12)(-9) + 6^2 \end{bmatrix} = \begin{bmatrix} 135 & -81 \\ -108 & 126 \end{bmatrix}$ Then $3A + 12A^2 = 3\begin{bmatrix} 3 & -9 \\ -12 & 6 \end{bmatrix} + 12\begin{bmatrix} 135 & -81 \\ -108 & 126 \end{bmatrix} = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$ Hence, $\operatorname{adj}(3A + 12A^2)$ = same matrix (since 2×2 case).

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