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

Given that $A \cdot \mathrm{adj}(A) = (\det A) I$ and $A \cdot \mathrm{adj}(A) = A A^T$, it follows that $(\det A) I = A A^T$.

For matrix $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$, the determinant is $\det A = (5a)(2) - (-b)(3) = 10a + 3b$.

Computing $A A^T$ yields:

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

Equating $(\det A) I = \begin{bmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{bmatrix}$ with $A A^T$ gives the system:

$15a - 2b = 0$

$10a + 3b = 13$

From $15a = 2b$, we have $b = \frac{15}{2}a$. Substituting into the second equation:

$10a + 3\left(\frac{15}{2}a\right) = 13 \Rightarrow 10a + \frac{45}{2}a = 13 \Rightarrow \frac{65}{2}a = 13 \Rightarrow a = \frac{2}{5}$

Then $b = \frac{15}{2} \cdot \frac{2}{5} = 3$.

These values satisfy the diagonal equation $10a + 3b = 25a^2 + b^2 = 13$, confirming consistency.

Therefore, $5a + b = 5\left(\frac{2}{5}\right) + 3 = 2 + 3 = 5$.

Final answer: $\boxed{5}$