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

\(\begin{aligned} & \mathrm{A}_{11}=(-1)^{1+1}(2)=2, \mathrm{~A}_{12}=(-1)^{1+2}(3)=-3 \\ & \mathrm{~A}_{21}=(-1)^{2+1}(-\mathrm{b})=\mathrm{b}, \mathrm{A}_{22}=(-1)^{2+2}(5 \mathrm{a})=5 \mathrm{a}\end{aligned}\)
\(\therefore \quad \operatorname{adj} A=\left[\begin{array}{cc}
2 & -3 \\
b & 5 a
\end{array}\right]^T=\left[\begin{array}{cc}
2 & b \\
-3 & 5 a
\end{array}\right]\)
Given, A adj \(\mathrm{A}=\mathrm{AA}^{\mathrm{T}}\)
\(\Rightarrow\left[\begin{array}{cc}5 a & - b \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}2 & b \\ -3 & 5 a \end{array}\right]=\left[\begin{array}{cc}5 a & - b \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}5 a & 3 \\ - b & 2\end{array}\right]\)
\(\Rightarrow\left[\begin{array}{cc}10 a +3 b & 0 \\ 0 & 3 b+10 a \end{array}\right]=\left[\begin{array}{cc}25 a ^2+ b ^2 & 15 a -2 b \\ 15 a -2 b & 13\end{array}\right]\)
\(\therefore \) by the equality of matrices,
\(15 a-2 b=0 \text { and } 3 b+10 a=13 \)
\( \Rightarrow a=\frac{2}{5} \text { and } b=3 \)
\( \therefore 5 a+b=5\left(\frac{2}{5}\right)+3=2+3=5\)