#### Provide RD Sharma solution for Maths Class 12 Chapter 5 Determinants Exercise Very short question Question 4 for maths textbook solution.

Hint: Here we use basic concept of singular matrix,

Let A is matrix then if $|A|=0$ A is singular

Given : $\left|\begin{array}{ll} 2 & 3 \\ 6 & 4 \end{array}\right|$

Solution : $|A|=\left|\begin{array}{ll} 2 & 3 \\ 6 & 4 \end{array}\right|$

$\rightarrow |A|=\left|\begin{array}{ll} 2 & 3 \\ 6 & 4 \end{array}\right|$

$\rightarrow$ Let's calculate the value of determinant

\begin{aligned} |A| &=\left|\begin{array}{ll} 2 & 3 \\ 6 & 4 \end{array}\right| \\ |A| &=2 \times 4-6 \times 3 \\ |A| &=8-18 \\ |A| &=(-10) \end{aligned}

$\rightarrow$ If $|A| \neq 0$ so, A is non singular matrix
$\rightarrow$ Because it doesnot satisfy the conditionn of singular matrix