#### Need RD Sharma solution for Maths Class 12 Chapter 5 Determinants Exercise Very short question Question 9 for maths textbook solution.

Hint: Here we use basic concept of matrix multiplication and determinant of  matrix

Given: $A=\left[\begin{array}{cc} 1 & 2 \\ 3 & -1 \end{array}\right] B=\left[\begin{array}{cc} 1 & 0 \\ -2 & 0 \end{array}\right] \text { Find }|A B|$

Solution :

For $\left | AB \right |$ let's Find AB
Here, A and B are $2 \times 2$ order of matrix.
So matrix multiplication is possible.

\begin{aligned} &A \times B=\left[\begin{array}{cc} 1 & 2 \\ 3 & -1 \end{array}\right] \times\left[\begin{array}{cc} 1 & 0 \\ -2 & 0 \end{array}\right] \\ &\; \; \; \; \; \; \; \; \; \; =\left[\begin{array}{cc} 1 \times 1+2(-2) & 0+0 \\ 3+2 & 0+0 \end{array}\right] \\ &\; \; \; \; \; A B=\left[\begin{array}{cc} -3 & 0 \\ 5 & 0 \end{array}\right] \end{aligned}

\begin{aligned} &\rightarrow \text { Let's Find }|A B|\\ &|A B|=[-3 \times 0]-[5 \times 0]\\ &\; \; \; \; \; \; \; =0-0\\ &|A B|=0 \end{aligned}