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
Answer & explanation
Correct answer: option 3
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: $\frac{65}{6}$