$\text{ If }A=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\text{ and }A\text{ adj }A=A{A}^T\text{, then }5a+b\text{ is equal to }$

$\underset{\_}{\underset{\_}{\text{ Given, }}}A=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\text{ and }\mathrm{A}\text{ adj }A=A{A}^T$

$\underset{\_}{\underset{\_}{\text{ Clearly }}}A(\mathrm{adj}A)=A\mid I_2$

$\left[\because \text{ if }\mathrm{A}\text{ is square matrix of order }\mathrm{n},\text{ then }A(\mathrm{adj}A)=(\mathrm{adj}A)\cdot A=A\mid I_n\right]$

$\begin{array}{l}A(\mathrm{adj}A)=\left|\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right|I_2\\ A(\mathrm{adj}A)=(10a+3b)I_2\\ A(\mathrm{adj}A)=(10a+3b)\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ A(\mathrm{adj}A)=\left[\begin{array}{cc}10a+3b& 0\\ 0& 10a+3b\end{array}\right]-----\left(1\right)\\ \text{ And }A{A}^T=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}5a& 3\\ -b& 2\end{array}\right]\end{array}$

$\begin{array}{l}A{A}^T=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]\\ \because A(\mathrm{adj}A)=A{A}^T\\ \therefore \left[\begin{array}{cc}10a+3b& 0\\ 0& 10a+3b\end{array}\right]=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]\end{array}$

[ using Eqs. (i) and (ii)]

$\begin{array}{r}\Rightarrow 15a-2b=0\\ \Rightarrow a=\frac{2b}{15}\cdots \cdots \cdots ...\left(2\right)\end{array}$

$\text{ And }10a+3b=13.---\left(3\right)$

On substituting the value of ‘a’ from Eq.(iii) in Eq.(iv), we get 

$\begin{array}{r}10\left(\frac{2b}{15}\right)+3b=13\\ \Rightarrow \frac{20b+45b}{15}=13\\ \Rightarrow \frac{65b}{15}=13\\ \Rightarrow b=3\end{array}$

Now , substituting the value of b in Eq. (iii) , we get $5a=2$ 
 Hence, 5a+b=2+3=5