Question

For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj} A) = \begin{bmatrix} 99 & 0 & 0 \\0 & 99 & 0 \\0 & 0 & 99 \end{bmatrix}$, then $\det(A)$ is equal to:

• A. \(3 \times 99\)
• B. \(99^3\)
• C. \(99^2\)
• D. 99

Hint:

The product \(A \cdot \operatorname{adj}(A)\) equals \(\det(A)\) times the identity matrix.

Answer 1

The Correct Option is D

Recall the matrix identity for any square matrix \(A\): \[ A \cdot \operatorname{adj}(A) = \det(A) I \] Where \(I\) is the identity matrix of the same order. Given: \[ A \cdot \operatorname{adj}(A) = \begin{bmatrix} 99 & 0 & 0 \\ 0 & 99 & 0 \\ 0 & 0 & 99 \end{bmatrix} = 99 I \] Comparing this with the identity: \[ A \cdot \operatorname{adj}(A) = \det(A) I \] We get: \[ \det(A) = 99 \]