Question

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \quad \text{and} \quad A \, \text{adj} \, A = A A^t, \quad \text{then} \, 5a + b \, \text{is equal to}$ 
 

• A. 5
• B. -1
• C. 4
• D. 13

Hint:

In matrix problems involving adjugates and transposition, use properties of determinants and the adjugate matrix. The determinant of \( A \) helps simplify the equations to solve for unknown variables.

Answer 1

The Correct Option is A

Given the matrix \( A \) as: \[ A = \begin{bmatrix} 5a & -b 3 & 2 \end{bmatrix} \] We are given the equation: \[ A \, \text{adj} \, A = A A^t \] We know that for any square matrix \( A \): \[ A \, \text{adj} \, A = \text{det}(A) \cdot I \] Where \( \text{det}(A) \) is the determinant of matrix \( A \) and \( I \) is the identity matrix. First, calculate the determinant of matrix \( A \): \[ \text{det}(A) = (5a)(2) - (3)(-b) = 10a + 3b \] Now, since the equation is \( A \, \text{adj} \, A = A A^t \), comparing both sides gives: \[ \text{det}(A) = \text{det}(A^t) \] Thus, the equation becomes: \[ 10a + 3b = 5a + b \] Simplifying: \[ 5a + 2b = 0 \] From this equation: \[ b = -\frac{5a}{2} \] Substitute this back into the expression \( 5a + b \): \[ 5a + b = 5a - \frac{5a}{2} = \frac{10a}{2} - \frac{5a}{2} = \frac{5a}{2} \] Now, equating the values: \[ 5a + b = 5 \] Thus, \( 5a + b = 5 \).