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

\(
\begin{aligned}
& \text { We have } \mathrm{A}\left[\begin{array}{cc}
5 \mathrm{a} & -\mathrm{b} \\
3 & 2
\end{array}\right] \text { and } \mathrm{A} \operatorname{adj} \mathrm{A}=\mathrm{AA}^{\mathrm{T}} \\
& \therefore\left[\begin{array}{cc}
5 \mathrm{a} & -\mathrm{b} \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & \mathrm{~b} \\
-3 & 5 \mathrm{a}
\end{array}\right]=\left[\begin{array}{cc}
5 \mathrm{a} & -\mathrm{b} \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
5 \mathrm{a} & 3 \\
-\mathrm{b} & 2
\end{array}\right] \\
& {\left[\begin{array}{cc}
10 \mathrm{a}+3 \mathrm{~b} & 0 \\
0 & 10 \mathrm{a}+3 \mathrm{~b}
\end{array}\right]=\left[\begin{array}{cc}
25 \mathrm{a}^2+\mathrm{b}^2 & 15 \mathrm{a}-2 \mathrm{~b} \\
15 \mathrm{a}-2 \mathrm{~b} & 9+4
\end{array}\right]} \\
& \therefore 10 \mathrm{a}+3 \mathrm{~b}=13 \text { and } 15 \mathrm{a}-2 \mathrm{~b}=0
\end{aligned}
\)
Solving these equations, we get \(a=\frac{2}{5}\) and \(b=3\)
\(
\therefore 5 a+b=2+3=5
\)