If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A\,(\text{adj }A) = AA^T$, Then $5a + b =$

Given $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$.

Using the property $A(\text{adj } A) = |A|I$, the given equation becomes $AA^T = |A|I$.

$|A| = (5a)(2) - (-b)(3) = 10a + 3b$

$AA^T = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \\ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{bmatrix}$

Equating $AA^T$ to $|A|I$:

$\begin{bmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{bmatrix} = \begin{bmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{bmatrix}$

From the off-diagonal elements:

$15a - 2b = 0 \Rightarrow b = \dfrac{15a}{2}$

From the bottom-right elements:

$10a + 3b = 13$

Substituting $b = \dfrac{15a}{2}$ into the equation:

$10a + 3\left(\dfrac{15a}{2}\right) = 13$

$\dfrac{20a + 45a}{2} = 13$

$\dfrac{65a}{2} = 13 \Rightarrow a = \dfrac{26}{65} = \dfrac{2}{5}$

Substituting $a = \dfrac{2}{5}$ to find $b$:

$b = \dfrac{15}{2} \left(\dfrac{2}{5}\right) = 3$

$5a + b = 5\left(\dfrac{2}{5}\right) + 3 = 2 + 3 = 5$

Answer: $5$