If $A =\begin{bmatrix}3&2a\\1&5\end{bmatrix}$ and $B =\begin{bmatrix}2&3\\b&5\end{bmatrix}$ both are singular matrices, then $a + b$ is equal to

$\frac{65}{6}$

The correct answer is Option (3) → $\frac{65}{6}$ **

Given:

$A = \begin{pmatrix} 3 & 2a \\ 1 & 5 \end{pmatrix}$ is singular.

For singular matrix, $\det(A) = 0$:

$\det(A) = 3\cdot 5 - (2a)(1) = 15 - 2a = 0$

$\Rightarrow 2a = 15 \Rightarrow a = \frac{15}{2}$

Similarly, $B = \begin{pmatrix} 2 & 3 \\ b & 5 \end{pmatrix}$ is singular.

$\det(B) = 2\cdot 5 - 3b = 10 - 3b = 0$

$\Rightarrow 3b = 10 \Rightarrow b = \frac{10}{3}$

Now compute $a + b$:

$a + b = \frac{15}{2} + \frac{10}{3}$

Common denominator 6:

$a + b = \frac{45}{6} + \frac{20}{6} = \frac{65}{6}$

Final Answer: $\frac{65}{6}$