#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 40 sub question (iii) math

Answer:  $x=\pm 4 \sqrt{3}$

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:

$\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$

Firstly,  we will multiply first two matrices

$\begin {array}{ll} \Rightarrow\left[\begin{array}{lll}x+0-2 & 0-10+0 & 2 x-5-3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0\\\\ \end{}$

$\begin {array}{ll} \Rightarrow[x(x-2)-40+2 x-8]=0\\\\ \Rightarrow x^{2}-2 x-40+2 x-8=0\\\\ \Rightarrow x^{2}-48=0\\\\ \Rightarrow x^{2}=48\\\\ \Rightarrow x=\pm \sqrt{48}\\\\ \Rightarrow x=\pm 4 \sqrt{3}\\\\ \end{}$