Question

If \( \det(AB) = \det(A)\det(B) \) and \( A \) is a non-singular matrix of order \( 3 \times 3 \), then \( \det(\text{adj}(A)) \) is:

• A. \( \det(A) \)
• B. \( (\det(A))^{-1} \)
• C. \( (\det(A))^2 \)
• D. \( (\det(A))^3 \)

Hint:

For any square matrix \( A \), the determinant of its adjugate matrix is given by \( \det(\text{adj}(A)) = (\det(A))^{n-1} \), where \( n \) is the order of the matrix.

Answer 1

The Correct Option is C

We are given that \( A \) is a non-singular matrix of order \( 3 \times 3 \), and we know the property: \[ \det(AB) = \det(A)\det(B) \] We are asked to find \( \det(\text{adj}(A)) \). For any square matrix \( A \), the determinant of its adjugate matrix is related to the determinant of \( A \) by the formula: \[ \det(\text{adj}(A)) = (\det(A))^{n-1} \] where \( n \) is the order of the matrix. Since \( A \) is a \( 3 \times 3 \) matrix, \( n = 3 \), so: \[ \det(\text{adj}(A)) = (\det(A))^{3-1} = (\det(A))^2 \] % Final Answer \[ \boxed{(\det(A))^2} \]

Answer 2

Step 1: Recall the properties of determinant and adjoint matrix.
Given \( A \) is a non-singular \( 3 \times 3 \) matrix.
We know that:
\[ \det(AB) = \det(A) \cdot \det(B). \]

Step 2: Use the relationship between \( A \) and its adjoint \( \text{adj}(A) \).
By definition:
\[ A \cdot \text{adj}(A) = (\det A) I, \] where \( I \) is the identity matrix.

Step 3: Take determinants on both sides.
\[ \det\big(A \cdot \text{adj}(A)\big) = \det\big((\det A) I\big). \]
Using the multiplicative property of determinants:
\[ \det(A) \cdot \det(\text{adj}(A)) = (\det A)^3, \] because \( \det\big(cI\big) = c^n \) for an \( n \times n \) identity matrix and scalar \( c \), here \( n=3 \).

Step 4: Solve for \( \det(\text{adj}(A)) \).
\[ \det(\text{adj}(A)) = \frac{(\det A)^3}{\det A} = (\det A)^2. \]

Hence, the determinant of the adjoint of \( A \) is \( (\det A)^2 \).