Question

If A = \(\begin{bmatrix}2a & 0& 0\\[0.3em]0& 2a& 0\\[0.3em]0&0 & 2a\\[0.3em] \end{bmatrix}\), then the value of \(|adj A|\)is:

• A. \(64a^6\)
• B. \(8a^3\)
• C. \(64a^3\)
• D. \(8a^6\)

Answer 1

The Correct Option is A

Given the matrix \( A = \begin{bmatrix} 2a & 0 & 0 \\ 0 & 2a & 0 \\ 0 & 0 & 2a \end{bmatrix} \), we need to find the value of \(|adj A|\).

The adjugate of a matrix, \( adj A \), is equal to the cofactor matrix of \( A \) transposed. For a \( 3 \times 3 \) matrix, if \( A \) is a diagonal matrix like the one given, every diagonal element \( 2a \) contributes to the determinant and adjugate in a specific pattern.

The determinant of a diagonal matrix is the product of its diagonal elements, so:

\[\det(A) = (2a) \times (2a) \times (2a) = 8a^3\]

For a \( 3 \times 3 \) matrix, the property \( adj(A) = \det(A) \cdot A^{-1} \) holds. Since \( A \) is a diagonal matrix, its inverse can be easily defined as:

\[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 2a & 0 & 0 \\ 0 & 2a & 0 \\ 0 & 0 & 2a \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2a} & 0 & 0 \\ 0 & \frac{1}{2a} & 0 \\ 0 & 0 & \frac{1}{2a} \end{bmatrix}\]

Hence, the adjugate is:

\[adj(A) = 8a^3 \begin{bmatrix} \frac{1}{2a} & 0 & 0 \\ 0 & \frac{1}{2a} & 0 \\ 0 & 0 & \frac{1}{2a} \end{bmatrix}\]

\[= \begin{bmatrix} 4a^2 & 0 & 0 \\ 0 & 4a^2 & 0 \\ 0 & 0 & 4a^2 \end{bmatrix}\]

The determinant of \( adj(A) \) is the product of its diagonal elements:

\[|adj(A)| = (4a^2) \times (4a^2) \times (4a^2) = 64a^6\]

Thus, \(|adj A| = 64a^6\).