#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 40 sub question (ii) math

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

First we multiply first two matrices,$\begin {array}{ll}\Rightarrow[1(1)+2(2)+(1)(1) \quad 1(2)+2(0)+1(0) \quad 1(0)+2(1)+1(2)]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{lll}1+4+1 & 2+0+0 & 0+2+2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0 \end{}$

$\begin {array}{ll}\\\\\Rightarrow\left[\begin{array}{lll}6 & 2 & 4\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0\\\\ \Rightarrow[0+2(2)+4(x)]=0\\\\ \Rightarrow 4+4 x=0\\\\ \Rightarrow 4 x=-4 \end{}$

$\begin {array}{ll}\\\\\Rightarrow x = -\frac{4}{4} = -1 \end{}$

Therefore, x=-1