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

Answer: $-70$

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

Given:$A=\left[\begin{array}{cc} 1 & 2 \\ 3 & -1 \end{array}\right], B=\left[\begin{array}{cc} 1 & -4 \\ 3 & -2 \end{array}\right]$

Solution :

To find $\left | AB \right |$, 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 & -4 \\ 3 & -2 \end{array}\right] \\ &=\left[\begin{array}{cc} 1 \times 1+2 \times 3 & 1(-4)+2(-2) \\ 3(1)+(-1)(3) & 3(-4)+(-1)(-2) \end{array}\right] \\ &A B=\left[\begin{array}{cc} 1+6 & -4-4 \\ 3-3 & -12+2 \end{array}\right] \\ &A B=\left[\begin{array}{cc} 7 & -8 \\ 0 & -10 \end{array}\right] \end{aligned}

$\rightarrow$ Let's find $\left | AB \right |$

\begin{aligned} &|A B|=7(-10)-(0)(-8) \\ &=(-70)-0 \\ &|A B|=-70 \end{aligned}