Question:
If $ A = \begin{pmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{pmatrix} $, then the value of $ \det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n $, then $ m + n $ is equal to:
If $ A = \begin{pmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{pmatrix} $, then the value of $ \det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n $, then $ m + n $ is equal to:
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To find determinants of adjugates, remember the formula \( \det(\text{adj}(M)) = \det(M)^{n-1} \) and apply it carefully.
Updated On: Apr 12, 2025
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The Correct Option is B
Solution and Explanation
We are given the matrix \( A \) and need to find the value of \( \det(\text{adj}(\text{adj}(3A))) \).
We know that the determinant of the adjugate of a matrix \( M \) is related to the determinant of \( M \) by the following formula: \[ \det(\text{adj}(M)) = \det(M)^{n-1} \] where \( n \) is the order of the matrix. We can calculate \( \det(3A) \), then use the formula to find \( \det(\text{adj}(\text{adj}(3A))) \).
After simplifying, we find that \( \det(\text{adj}(\text{adj}(3A))) = 2^{24} \cdot 3^{24} \), so \( m + n = 24 \).
We know that the determinant of the adjugate of a matrix \( M \) is related to the determinant of \( M \) by the following formula: \[ \det(\text{adj}(M)) = \det(M)^{n-1} \] where \( n \) is the order of the matrix. We can calculate \( \det(3A) \), then use the formula to find \( \det(\text{adj}(\text{adj}(3A))) \).
After simplifying, we find that \( \det(\text{adj}(\text{adj}(3A))) = 2^{24} \cdot 3^{24} \), so \( m + n = 24 \).
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