Question

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

Answer 1

5

Answer 2

Given:
\(A = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}\)

We need to find \(\text{adj}A\):
\(\text{adj}A = \begin{bmatrix} 2 & 3b \\ -3 & 2a \end{bmatrix}\)

Now, we compute \(A \cdot \text{adj}A\):
\(A \cdot \text{adj}A = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3b \\ -3 & 2a \end{bmatrix}\)

\(= \begin{bmatrix} (2a \cdot 2) + (-3b \cdot -3) & (2a \cdot 3b) + (-3b \cdot 2a) \\ (3 \cdot 2) + (2 \cdot -3) & (3 \cdot 3b) + (2 \cdot 2a) \end{bmatrix}\)

\(= \begin{bmatrix} 4a + 9b & 0 \\ 0 & 9b + 4a \end{bmatrix}\)

We also compute \(A \cdot A^T\):
\(A^T = \begin{bmatrix} 2a & 3 \\ -3b & 2 \end{bmatrix}\)

\(A \cdot A^T = \begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2a & 3 \\ -3b & 2 \end{bmatrix}\)

\(= \begin{bmatrix} (2a \cdot 2a) + (-3b \cdot -3b) & (2a \cdot 3) + (-3b \cdot 2) \\ (3 \cdot 2a) + (2 \cdot -3b) & (3 \cdot 3) + (2 \cdot 2) \end{bmatrix}\)

\(= \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 9 + 4 \end{bmatrix}\)

\(= \begin{bmatrix} 4a^2 + 9b^2 & 6a - 6b \\ 6a - 6b & 13 \end{bmatrix}\)

Given that \(\text{adj}A = A \cdot A^T\), we equate:
\(\begin{bmatrix} 4a + 9b & 0 \\ 0 & 4a + 9b \end{bmatrix} = \begin{bmatrix} 4a^2 + 9b^2 \\ 6a - 6b \end{bmatrix}\)

From this, we get the following system of equations:
1. \(4a + 9b = 4a^2 + 9b^2\)
2. \(6a - 6b = 0\) (from the off-diagonal elements)

From equation 2:
\(6a - 6b = 0 \implies a = b\)

Substitute a = b into equation 1:
\(4a + 9a = 4a^2 + 9a^2\)
\(13a = 13a^2\)
\(a^2 = a\)
\(a(a - 1) = 0\)

So, \(a = 0\) or \(a = 1\). Since \(a = b\), we have:
\(a = b = 1\)
\(2a + 3b = 2(1) + 3(1) = 2 + 3 = 5\)

So, the answer is 5.