Question

If $A$ is a square matrix of order 3 and $A \cdot (\operatorname{Adj}(A)) = 10I$, then the value of $\frac{1}{25} |\operatorname{adj}(A)|$ is:

• A. 100
• B. 25
• C. 10
• D. 4

Hint:

\(|\operatorname{adj}(A)| = |\det(A)|^{n-1}\) for an \(n \times n\) matrix \(A\).

Answer 1

The Correct Option is A

Step 1: Using the identity: \[ A \cdot \operatorname{adj}(A) = \det(A) I \] Given: \[ A \cdot \operatorname{adj}(A) = 10 I \] So, \[ \det(A) = 10 \] Step 2: For a square matrix of order \(n\), the determinant of the adjoint matrix is related to the determinant of the original matrix as: \[ |\operatorname{adj}(A)| = |\det(A)|^{n-1} \] Since \(n = 3\): \[ |\operatorname{adj}(A)| = |\det(A)|^{2} = 10^2 = 100 \] Step 3:  Given \(A \cdot \operatorname{adj}(A) = 10 I\), so \(\det(A) = 10\). 
Therefore, \[ |\operatorname{adj}(A)| = 10^{2} = 100 \] Then, \[ \frac{1}{25} |\operatorname{adj}(A)| = \frac{100}{25} = 4 \] But the correct answer in options is 4, option (d). So the correct answer is 4.