Question

If \(A=\begin{bmatrix} 2a & -3b\\ 3 & 2\end{bmatrix}\) and \(\text{adj}A = AA^T\), then 2a+3b is?

Answer 1

Given:
\[ A = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix} \] We need to find the adjugate of \( A \), denoted by \(\text{adj}A\): \[ \text{adj}A = \begin{bmatrix} 2 & 3b \\ -3 & 2a \end{bmatrix} \]

Step 1: Compute \( A \cdot \text{adj}A \):
\[ A \cdot \text{adj}A = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3b \\ -3 & 2a \end{bmatrix} \] \[ = \begin{bmatrix} 4a + 9b & 0 \\ 0 & 9b + 4a \end{bmatrix} \]

Step 2: Compute \( A \cdot A^T \):
\[ A^T = \begin{bmatrix} 2a & 3 \\ -3b & 2 \end{bmatrix} \] \[ A \cdot A^T = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2a & 3 \\ -3b & 2 \end{bmatrix} \] \[ = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 13 \end{bmatrix} \]

Step 3: Equate \( \text{adj}A \) and \( A \cdot A^T \):
Since \( \text{adj}A = A \cdot A^T \), we have: \[ \begin{bmatrix} 4a + 9b & 0 \\ 0 & 4a + 9b \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 13 \end{bmatrix} \] This gives the system of equations:
1. \( 4a + 9b = 4a^2 + 9b^2 \)
2. \( 6a - 6b = 0 \) (from the off-diagonal elements)

Step 4: Solve the system:
From equation 2: \( 6a - 6b = 0 \implies a = b \). Substitute \( a = b \) into equation 1: \[ 4a + 9a = 4a^2 + 9a^2 \implies 13a = 13a^2 \implies a^2 = a \] \[ a(a - 1) = 0 \implies a = 0 \text{ or } a = 1 \] Since \( a = b \), we have \( a = b = 1 \).

Step 5: Find the final answer:
\[ 2a + 3b = 2(1) + 3(1) = 2 + 3 = 5 \] Therefore, the answer is \( \boxed{5} \).