If $A$ is a square matrix of order 3 such that $|A| = 2$, then $|\operatorname{adj}(\operatorname{adj}(A))|$ is:

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

Step 1: The determinant of the adjugate of an $n \times n$ matrix $A$ is given by $|\operatorname{adj}(A)| = |\det(A)|^{n-1}$. For $n = 3$ and $|A| = 2$, we have $|\operatorname{adj}(A)| = 2^{3-1} = 2^{2} = 4$. Step 2: To find $|\operatorname{adj}(\operatorname{adj}(A))|$, we treat $\operatorname{adj}(A)$ as a new $3 \times 3$ matrix. Applying the same formula, $|\operatorname{adj}(\operatorname{adj}(A))| = |\det(\operatorname{adj}(A))|^{3-1} = |\det(\operatorname{adj}(A))|^{2}$. Step 3: We know that $\det(\operatorname{adj}(A)) = |\operatorname{adj}(A)|$. From Step 1, $\det(\operatorname{adj}(A)) = 4$. Step 4: Substituting the value from Step 3 into the equation from Step 2, we get $|\operatorname{adj}(\operatorname{adj}(A))| = 4^{2} = 16$.

If $A$ is a square matrix of order 3 such that $|A| = 2$, then $|\operatorname{adj}(\operatorname{adj}(A))|$ is:

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The Correct Option is C

Solution and Explanation

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