#### Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 30 maths textbook solution

VSQ : 30

Answer : $\left[\begin{array}{ll} 2 & 3 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 17 & -7 \end{array}\right]$

Given : $\left[\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 9 & -4 \end{array}\right]$

and elementary operation $R_{2}\rightarrow R_{2}+R_{1}$

Hint : In multiplication row operation is equivalent to left multiplication

Solution :

$\left[\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 9 & -4 \end{array}\right]$

By applying operation $R_{2}\rightarrow R_{2}+R_{1}$

$\left[\begin{array}{ll} 2 & 3 \\ 3 & 7 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 8 & -3 \\ 17 & -7 \end{array}\right]$