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

Answer: (D)

216

Answer 2

Given that $B$ is the matrix of cofactors of $A$, we use the relationship:

$AB = \det(A) \cdot I_3,$ where $I_3$ is the $3 \times 3$ identity matrix. Therefore: $\det(AB) = \det(A)^3.$

Step 1: Equating the Cofactor Condition

We know: $(2\alpha^2 - 3\alpha) = \alpha.$

Rearranging: $2\alpha^2 - 3\alpha - \alpha = 0 \implies 2\alpha^2 - 4\alpha = 0.$

Since $\alpha \neq 0$, we get: $\alpha = 2.$

Step 2: Substitute and Find $\beta$

Using the relation: $2\alpha^2 - \alpha\beta = 3\alpha,$ substitute $\alpha = 2$: $2 \cdot 2^2 - 2\beta = 3 \cdot 2 \implies 8 - 2\beta = 6 \implies 2\beta = 2 \implies \beta = 1.$

Step 3: Calculate $\det(A)$

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 of $A$ is: $\det(A) = 1 \cdot \begin{vmatrix} 2 & 1 \\\ 2 & 4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\\ -1 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 2 \\\ -1 & 2 \end{vmatrix}.$

Calculating each minor: $\begin{vmatrix} 2 & 1 \\\ 2 & 4 \end{vmatrix} = 2 \cdot 4 - 1 \cdot 2 = 6,$ $\begin{vmatrix} 2 & 1 \\\ -1 & 4 \end{vmatrix} = 2 \cdot 4 - 1 \cdot (-1) = 9,$ $\begin{vmatrix} 2 & 2 \\\ -1 & 2 \end{vmatrix} = 2 \cdot 2 - 2 \cdot (-1) = 6.$

Thus: $\det(A) = 1 \cdot 6 - 2 \cdot 9 + 3 \cdot 6 = 6 - 18 + 18 = 6.$

Step 4: Calculate $\det(AB)$

Since: $\det(AB) = \det(A)^3 = 6^3 = 216.$

Therefore, the correct answer is Option (4).