Question

If \( B = \begin{bmatrix} 1 & 3 \\ 2 & \alpha \end{bmatrix} \) is the adjoint of a matrix \( A \) and \( |A| = 2 \), then the value of \( \alpha \) is:

• A. 5
• B. 2
• C. 3
• D. 4

Hint:

For adjoint matrices, remember the relationship \( A \cdot B = |A| I \). This can help solve for unknowns like \( \alpha \).

Answer 1

The Correct Option is A

If \( B = \begin{bmatrix} 1 & 3 \\ 2 & \alpha \end{bmatrix} \) is the adjoint of a matrix \( A \) and \( |A| = 2 \), then the value of \( \alpha \) is:

Step 1: Recall the properties of adjoint matrices
The adjoint of a matrix \( A \), denoted as \( \text{adj}(A) \), is the transpose of the cofactor matrix of \( A \). If \( A \) is a 2x2 matrix, then: \[ \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] where \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Thus, we have the following relationships: \[ B = \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

Step 2: Compare the matrices
From the problem statement, we know: \[ B = \begin{bmatrix} 1 & 3 \\ 2 & \alpha \end{bmatrix} \] By comparing with the general form of \( \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \), we can deduce: \[ d = 1, \quad -b = 3, \quad -c = 2, \quad a = \alpha \] Thus, we have: \[ b = -3, \quad c = -2, \quad a = \alpha \]

Step 3: Use the property \( A \times \text{adj}(A) = |A| \cdot I \)
The property of the adjoint matrix states that: \[ A \times \text{adj}(A) = |A| \cdot I \] where \( I \) is the identity matrix and \( |A| \) is the determinant of \( A \). Since \( |A| = 2 \), we have: \[ A \times \text{adj}(A) = 2 \cdot I \]

Step 4: Write the matrix \( A \) and calculate the product
Let \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} \alpha & -3 \\ -2 & 1 \end{bmatrix} \). Now, compute the product \( A \times \text{adj}(A) \): \[ A \times \text{adj}(A) = \begin{bmatrix} \alpha & -3 \\ -2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 2 & \alpha \end{bmatrix} \] Multiplying the matrices: \[ A \times \text{adj}(A) = \begin{bmatrix} \alpha \cdot 1 + (-3) \cdot 2 & \alpha \cdot 3 + (-3) \cdot \alpha \\ (-2) \cdot 1 + 1 \cdot 2 & (-2) \cdot 3 + 1 \cdot \alpha \end{bmatrix} \] Simplifying the entries: \[ A \times \text{adj}(A) = \begin{bmatrix} \alpha - 6 & 0 \\ 0 & -6 + \alpha \end{bmatrix} \]

Step 5: Set the product equal to \( 2I \)
From the property \( A \times \text{adj}(A) = 2 \cdot I \), we know that: \[ A \times \text{adj}(A) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \] Therefore, equating the two matrices: \[ \begin{bmatrix} \alpha - 6 & 0 \\ 0 & \alpha - 6 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \] From this, we get: \[ \alpha - 6 = 2 \] Solving for \( \alpha \): \[ \alpha = 8 \]

Conclusion:
The value of \( \alpha \) is \( \boxed{8} \).