Question

If \( A = \begin{bmatrix} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{bmatrix} \), then \( \sqrt{|\text{Adj } A|} = \)

• A. 64
• B. 16
• C. 36
• D. 216

Hint:

Remember the identity \( |\text{Adj } A| = |A|^{n-1} \). For a \( 3 \times 3 \) matrix, \( \sqrt{|\text{Adj } A|} \) is simply the absolute value of the determinant of A.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We need to find the square root of the determinant of the adjoint of matrix \( A \). The property relating \( |\text{Adj } A| \) to \( |A| \) is \( |\text{Adj } A| = |A|^{n-1} \), where \( n \) is the order of the matrix.
Step 2: Key Formula or Approach:

Here, \( n = 3 \). So, \( |\text{Adj } A| = |A|^{3-1} = |A|^2 \). Therefore, \( \sqrt{|\text{Adj } A|} = \sqrt{|A|^2} = | |A| | \).
Step 3: Detailed Explanation:

Calculate \( |A| \): \[ |A| = 1 \begin{vmatrix} 4 & -6 \\ -11 & 13 \end{vmatrix} - (-3) \begin{vmatrix} -2 & -6 \\ 7 & 13 \end{vmatrix} + (-5) \begin{vmatrix} -2 & 4 \\ 7 & -11 \end{vmatrix} \] \[ |A| = 1(52 - 66) + 3(-26 - (-42)) - 5(22 - 28) \] \[ |A| = 1(-14) + 3(-26 + 42) - 5(-6) \] \[ |A| = -14 + 3(16) + 30 \] \[ |A| = -14 + 48 + 30 = 64 \] Since \( |A| = 64 \), we have: \[ \sqrt{|\text{Adj } A|} = \sqrt{64^2} = 64 \]
Step 4: Final Answer:

The value is 64.