#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 40 sub question (iv) maths textbook solution

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:

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

Firstly, we will multiply first two matrices

$\begin {array} {ll} \Rightarrow[2 x(1)+3(-3) \quad 2 x(2)+3(0)]\left[\begin{array}{l}x \\ 8\end{array}\right]=0\\\\ \Rightarrow\left[\begin{array}{ll}2 x-9 & 4 x+0\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0\\\\ \end {}$

$\begin {array} {ll}\Rightarrow\left[\begin{array}{ll}2 x-9 & 4 x\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0\\\\ \Rightarrow[(2 x-9) x+4 x(8)]=0\\\\ \Rightarrow[(2 x-9) x+4 x(8)]=0\\\\ \Rightarrow 2 x^{2}-9 x+32 x=0\\\\ \end {}$

$\begin{array}{l} \Rightarrow 2 x^{2}+23 x=0 \\\\ \Rightarrow x(2 x+23)=0 \\\\ \Rightarrow x=0 \quad \text { or } \quad 2 x+23=0 \\\\\end{array}$

$\begin{array}{l} 2 x=-23 \\\\ x=-\frac{23}{2} \\\\\end{array}$