Question

If for any \( 2 \times 2 \) square matrix \( A \), \( A(\text{adj} A) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \), then find the value of \( \det(A) \).

• A. 6
• B. 7
• C. 8
• D. 5

Hint:

The product of a matrix and its adjugate results in the determinant of the matrix multiplied by the identity matrix. This property is useful for solving determinant-related problems.

Answer 1

The Correct Option is C

Step 1: Use the property of adjugates

For any square matrix \( A \), it is known that: \[ A \cdot \text{adj}(A) = \det(A) \cdot I \] where \( \text{adj}(A) \) is the adjugate of \( A \) and \( I \) is the identity matrix. In this case, we are given: \[ A \cdot \text{adj}(A) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \] This is equivalent to: \[ A \cdot \text{adj}(A) = 8 \cdot I \] where \( I \) is the identity matrix.

Step 2: Compare both equations

Using the property of adjugates, we can equate the two expressions: \[ \det(A) \cdot I = 8 \cdot I \] This implies: \[ \det(A) = 8 \]

Final Answer:

The value of \( \det(A) \) is \( \boxed{8} \).