Question

Let \( A \) be a \( 4 \times 4 \) matrix and \( P \) be its adjoint matrix. If \( |P| = \left| \frac{A}{2} \right| \), then \( |A^{-1}| = ? \)

• A. \( \pm \frac{1}{4} \)
• B. \( \pm 8 \)
• C. \( \pm 2 \)
• D. \( \pm 4 \)

Hint:

For an \( n \times n \) matrix \( A \), the determinant of the adjoint matrix \( P \) is given by \( |P| = |A|^{n-1} \). Use this to solve determinant-related problems.

Answer 1

The Correct Option is D

We are given that \( A \) is a \( 4 \times 4 \) matrix and \( P \) is its adjoint matrix. The determinant of the adjoint matrix \( P \) is given by: \[ |P| = \left| \frac{A}{2} \right|. \] We know that for a matrix \( A \), the relation between \( |A| \) and \( |P| \) is given by: \[ |P| = |A|^{n-1}, \] where \( n \) is the order of the matrix \( A \). Since \( A \) is a \( 4 \times 4 \) matrix, \( n = 4 \). Thus, we have: \[ |P| = |A|^{4-1} = |A|^3. \] Now, we are given \( |P| = \left| \frac{A}{2} \right| \). The determinant of \( \frac{A}{2} \) is: \[ \left| \frac{A}{2} \right| = \frac{|A|}{2^4} = \frac{|A|}{16}. \] Thus, we have the equation: \[ |A|^3 = \frac{|A|}{16}. \] Solving for \( |A| \), we get: \[ |A|^3 = \frac{|A|}{16} \quad \Rightarrow \quad |A|^2 = \frac{1}{16} \quad \Rightarrow \quad |A| = \pm \frac{1}{4}. \] Finally, the determinant of \( A^{-1} \) is: \[ |A^{-1}| = \frac{1}{|A|} = \pm 4. \] Thus, \( |A^{-1}| = \pm 4 \).