If $$A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$$ and $$A \cdot adj A = A A^T$$, then $$5a + b$$ is equal to

We have the square matrix

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

For a $$2\times2$$ matrix $$ \begin{bmatrix} p & q \\ r & s \end{bmatrix}, $$ the adjugate (co-factor transpose) is given by the formula

$$ \operatorname{adj}A=\begin{bmatrix} s & -q \\ -r & p \end{bmatrix}. $$

Comparing, we identify

$$ p=5a,\; q=-b,\; r=3,\; s=2, $$

so that

$$ \operatorname{adj}A =\begin{bmatrix} 2 & -(-b)\\ -3 & 5a \end{bmatrix} =\begin{bmatrix} 2 & b\\ -3 & 5a \end{bmatrix}. $$

Now we compute the product $$A\operatorname{adj}A$$ :

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

Multiplying row by column, step by step,

$$ \begin{aligned} (1,1)&: 5a\cdot2 + (-b)\cdot(-3) = 10a + 3b,\\[2mm] (1,2)&: 5a\cdot b + (-b)\cdot 5a = 5ab - 5ab = 0,\\[2mm] (2,1)&: 3\cdot2 + 2\cdot(-3) = 6 - 6 = 0,\\[2mm] (2,2)&: 3\cdot b + 2\cdot 5a = 3b + 10a. \end{aligned} $$

Hence

$$ A\operatorname{adj}A =\begin{bmatrix} 10a+3b & 0\\ 0 & 3b+10a \end{bmatrix}. $$

Next we need $$A A^{T}$$. First, the transpose of $$A$$ is

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

Now multiply $$A$$ by $$A^{T}$$ :

$$ AA^{T} =\begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \\ -b & 2 \end{bmatrix}. $$

Again, compute every entry explicitly:

$$ \begin{aligned} (1,1)&: 5a\cdot5a + (-b)\cdot(-b) = 25a^{2} + b^{2},\\[2mm] (1,2)&: 5a\cdot3 + (-b)\cdot2 = 15a - 2b,\\[2mm] (2,1)&: 3\cdot5a + 2\cdot(-b) = 15a - 2b,\\[2mm] (2,2)&: 3\cdot3 + 2\cdot2 = 9 + 4 = 13. \end{aligned} $$

Therefore

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

The condition given in the question is

$$ A\operatorname{adj}A = AA^{T}. $$

Equating corresponding entries gives three independent equations:

$$ \begin{aligned} 10a + 3b &= 25a^{2} + b^{2}, \quad -(1)\\ 0 &= 15a - 2b, \quad -(2)\\ 3b + 10a &= 13. \quad -(3) \end{aligned} $$

From equation (2) we get

$$ 15a - 2b = 0 \;\Longrightarrow\; 2b = 15a \;\Longrightarrow\; b = \frac{15}{2}\,a. $$

Substituting this value of $$b$$ into equation (3):

$$ 3\Bigl(\frac{15}{2}a\Bigr) + 10a = 13 \;\Longrightarrow\; \frac{45}{2}a + 10a = 13 \;\Longrightarrow\; \frac{45a + 20a}{2} = 13 \;\Longrightarrow\; \frac{65a}{2} = 13. $$

Solving for $$a$$,

$$ a = 13 \times \frac{2}{65} = \frac{26}{65} = \frac{2}{5}. $$

Now substitute $$a=\dfrac{2}{5}$$ back into $$b=\dfrac{15}{2}a$$ :

$$ b = \frac{15}{2}\times\frac{2}{5} = \frac{30}{10} = 3. $$

Finally, we compute the required expression $$5a + b$$ :

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

All three equations (1), (2) and (3) are satisfied by these values, confirming the correctness of the solution. Therefore

$$ 5a + b = 5. $$ Hence, the correct answer is Option D.