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. 280
• D. 174

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

Step 1: Recall important properties.
For any square matrix \( A \) of order \( n \): \[ \text{adj}(A) = |A| A^{-1}, \quad \text{and} \quad |\text{adj}(A)| = |A|^{n-1}. \] Also, \( \text{adj}(kA) = k^{n-1}\text{adj}(A) \) for scalar \( k \). 

Since \( A \) is \( 3 \times 3 \), we have \( n = 3 \).

Step 2: Compute \( \text{adj}(2A) \).
\[ \text{adj}(2A) = 2^{n-1}\text{adj}(A) = 2^2 \text{adj}(A) = 4 \text{adj}(A). \]

Step 3: Now compute \( B = \text{adj}(\text{adj}(2A)) \).
Let \( C = \text{adj}(2A) = 4 \text{adj}(A) \). Then \( B = \text{adj}(C) = \text{adj}(4 \text{adj}(A)) \). Using the property \( \text{adj}(kM) = k^{n-1}\text{adj}(M) \), we get: \[ B = 4^{2} \text{adj}(\text{adj}(A)) = 16 \, \text{adj}(\text{adj}(A)). \]

Step 4: Simplify \( \text{adj}(\text{adj}(A)) \).
For a \( 3 \times 3 \) matrix, the formula is: \[ \text{adj}(\text{adj}(A)) = |A|^{n-2} A = |A|^{1} A = |A|A. \] So: \[ B = 16(|A|A) = 16|A|A. \]

Step 5: Substitute \( |A| = \frac{1}{2} \).
\[ B = 16 \times \frac{1}{2} A = 8A. \]

Step 6: Compute \( |B| \) and \( \text{trace}(B) \).
For a scalar multiple of a matrix: \[ |kA| = k^n |A|, \quad \text{trace}(kA) = k \cdot \text{trace}(A). \] Thus: \[ |B| = |8A| = 8^3 |A| = 512 \times \frac{1}{2} = 256, \] \[ \text{trace}(B) = 8 \times \text{trace}(A) = 8 \times 3 = 24. \]

Step 7: Compute the required sum.
\[ |B| + \text{trace}(B) = 256 + 24 = 280. \]

Final Answer:

\[ \boxed{280} \]

Answer 2

For a $3\times3$ matrix $A$ we have $\text{adj}(kA)=k^{2}\text{adj}(A)$ and $\text{adj}(\text{adj}(A))=|A|A$. With $|A|=\tfrac12$ and $\operatorname{tr}(A)=3$: 

\[ \text{adj}(2A)=2^{2}\text{adj}(A)=4\text{adj}(A). \] Hence \[ B=\text{adj}(\text{adj}(2A))=\text{adj}(4\text{adj}(A))=4^{2}\,\text{adj}(\text{adj}(A)) =16\cdot(|A|A)=16\cdot\frac12\,A=8A. \]

Therefore \[ |B|=|8A|=8^{3}|A|=512\cdot\frac12=256,\qquad \operatorname{tr}(B)=8\operatorname{tr}(A)=8\cdot3=24. \]

Final result: \[ |B|+\operatorname{tr}(B)=256+24=\boxed{280}. \]