#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (iii) math

$A=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$

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:

$\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$

We know that two matrices B and C are eligible for the product BC only when number of columns of B is equal to number or rows of C.

So, from the given definition we can consider that the order of matrix A is $1 \times 3$  i.e. we can assume $\therefore\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]_{3 \times 1}\left[\begin{array}{lll}x_{1} & x_{2} & x_{3}\end{array}\right]_{1 \times 3}=\left[\begin{array}{ccc}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]_{3 \times 3}$

$\\\\ \Rightarrow\left[\begin{array}{lll} 4\left(x_{1}\right) & 4\left(x_{2}\right) & 4\left(x_{3}\right) \\ 1\left(x_{1}\right) & 1\left(x_{2}\right) & 1\left(x_{3}\right) \\ 3\left(x_{1}\right) & 3\left(x_{2}\right) & 3\left(x_{3}\right) \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} 4 x_{1} & 4 x_{2} & 4 x_{3} \\ x_{1} & x_{2} & x_{3} \\ 3 x_{1} & 3 x_{2} & 3 x_{3} \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]$

Equating the corresponding element of the two matrices, we have

$x_{1}=-1, x_{2}=2, x_{3}=1$

So, matrix  $A=\left[\begin{array}{lll} -1 & 2 & 1\end{array}]\right.$