For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj} A) = \begin{bmatrix} 99 & 0 & 0 \\0 & 99 & 0 \\0 & 0 & 99 \end{bmatrix}$, then $\det(A)$ is equal to:

Question

For the matrices $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} $, if $ (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $, then among the following which one is true?}

Answer 1

For any square matrix $A$, the matrix identity is $A \cdot \operatorname{adj}(A) = \det(A) I$, where $I$ is the identity matrix of the same order. We are given that $A \cdot \operatorname{adj}(A) = \begin{bmatrix} 99 & 0 & 0 \\ 0 & 99 & 0 \\ 0 & 0 & 99 \end{bmatrix} = 99 I$. Comparing this with the identity $A \cdot \operatorname{adj}(A) = \det(A) I$, we deduce that $\det(A) = 99$.

For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj} A) = \begin{bmatrix} 99 & 0 & 0 \\0 & 99 & 0 \\0 & 0 & 99 \end{bmatrix}$, then $\det(A)$ is equal to:

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The Correct Option is D

Solution and Explanation

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