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

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Answer: Non singular matrix

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

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