Question

Let $\alpha \beta \neq 0$ and $A = \begin{bmatrix} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{bmatrix}$. If $B = \begin{bmatrix} 3\alpha & -9 & 3\alpha \\ -\alpha & 7 & -2\alpha \\ -2\alpha & 5 & -2\beta \end{bmatrix}$ is the matrix of cofactors of the elements of A, then det(AB) is equal to:

• A. 343
• B. 125
• C. 64
• D. 216

Answer 1

The Correct Option is D

Given that \( B \) is the matrix of cofactors of \( A \), we apply the relation \( AB = \det(A) \cdot I_3 \), where \( I_3 \) is the \( 3 \times 3 \) identity matrix. This implies \( \det(AB) = \det(A)^3 \).

Step 1: Determine \( \alpha \)

The cofactor condition is \( (2\alpha^2 - 3\alpha) = \alpha \). Rearranging yields \( 2\alpha^2 - 4\alpha = 0 \). Since \( \alpha eq 0 \), we find \( \alpha = 2 \).

Step 2: Determine \( \beta \)

Using \( 2\alpha^2 - \alpha\beta = 3\alpha \) and substituting \( \alpha = 2 \): \( 2 \cdot 2^2 - 2\beta = 3 \cdot 2 \). Simplifying gives \( 8 - 2\beta = 6 \), which leads to \( 2\beta = 2 \) and \( \beta = 1 \).

Step 3: Compute \( \det(A) \)

Substitute \( \alpha = 2 \) and \( \beta = 1 \) into matrix \( A \): \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{bmatrix}. \]

The determinant is calculated as: \[ \det(A) = 1 \cdot \begin{vmatrix} 2 & 1 \\ 2 & 4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ -1 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 2 \\ -1 & 2 \end{vmatrix}. \]

The minors evaluate to: \( \begin{vmatrix} 2 & 1 \\ 2 & 4 \end{vmatrix} = 6 \), \( \begin{vmatrix} 2 & 1 \\ -1 & 4 \end{vmatrix} = 9 \), and \( \begin{vmatrix} 2 & 2 \\ -1 & 2 \end{vmatrix} = 6 \).

Therefore, \( \det(A) = 1 \cdot 6 - 2 \cdot 9 + 3 \cdot 6 = 6 - 18 + 18 = 6 \).

Step 4: Compute \( \det(AB) \)

Using the relation \( \det(AB) = \det(A)^3 \): \( \det(AB) = 6^3 = 216 \).

The solution corresponds to Option (4).