Question

For a \( 3 \times 3 \) matrix \( M \), let trace(M) denote the sum of all the diagonal elements of \( M \). Let \( A \) be a \( 3 \times 3 \) matrix such that \( |A| = \frac{1}{2} \) and \( \text{trace}(A) = 3 \). If \( B = \text{adj}(\text{adj}(2A)) \), then the value of \( |B| + \text{trace}(B) \) equals:

• A. 132
• B. 56
• C. 174
• D. 280

Hint:

When working with determinants and traces of adjugate matrices: - Use the property \( \text{adj}(M) = |M|^{n-1} M^{-1} \) to compute the determinant of the adjugate. - For the trace of the adjugate matrix, use the relation \( \text{trace}(\text{adj}(A)) = (n-1) \cdot \text{trace}(A) \) to simplify calculations.

Answer 1

The Correct Option is C

Given a \( 3 \times 3 \) matrix \( A \) with \( |A| = \frac{1}{2} \) and \( \text{trace}(A) = 3 \). The adjugate of a matrix \( M \), denoted \( \text{adj}(M) \), satisfies \( M \cdot \text{adj}(M) = |M| I \). We are given \( B = \text{adj}(\text{adj}(2A)) \). Using the property \( \text{adj}(M) = |M|^{n-1} M^{-1} \) with \( n=3 \), we first determine \( |B| \). The determinant of the adjugate is \( | \text{adj}(M) | = |M|^{n-1} \). Therefore, \( | \text{adj}(2A) | = |2A|^2 \). Since \( |2A| = 2^3 |A| = 8 \cdot \frac{1}{2} = 4 \), we have \( | \text{adj}(2A) | = 4^2 = 16 \). Consequently, \( |B| = | \text{adj}(\text{adj}(2A)) | = | \text{adj}(2A) |^{3-1} = 16^2 = 256 \). Next, we compute \( \text{trace}(B) \). The trace of the adjugate is related to the trace of the original matrix by \( \text{trace}(\text{adj}(A)) = (n-1) \cdot \text{trace}(A) \). For \( \text{adj}(2A) \), \( \text{trace}(\text{adj}(2A)) = (3-1) \cdot \text{trace}(2A) = 2 \cdot (2 \cdot \text{trace}(A)) = 4 \cdot 3 = 12 \). Thus, \( \text{trace}(B) = \text{trace}(\text{adj}(\text{adj}(2A))) \). Using \( M = \text{adj}(2A) \), we have \( \text{trace}(\text{adj}(M)) = (n-1) \text{trace}(M) \). Here \( M = \text{adj}(2A) \) is a \( 3 \times 3 \) matrix. So, \( \text{trace}(B) = \text{trace}(\text{adj}(\text{adj}(2A))) = (3-1) \cdot \text{trace}(\text{adj}(2A)) = 2 \cdot 12 = 24 \). The final computation is \( |B| + \text{trace}(B) = 256 + 24 = 280 \).