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 |Adj A| to |A| is:

|Adj A| = |A|n-1

where n is the order of the matrix.

Step 2: Key Formula or Approach:

Here, n = 3. So,

|Adj A| = |A|3-1 = |A|2

Therefore,

√|Adj A| = √(|A|2) = ||A||

Step 3: Detailed Explanation:

Calculate |A|:

|A| = 1 × 

4	-6
-11	13

- (-3) × 

-2	-6
7	13

+ (-5) × 

-2	4
7	-11

|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:

√|Adj A| = √(642) = 64

Step 4: Final Answer:

The value is 64.