#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 49

$A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$

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:

$A\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6 I_{2}$

is identity matrix of order 2

$I_{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Now, let

$A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$\\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ll}a+b & -2 a+4 b \\ c+d & -2 c+4 d\end{array}\right]=\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$

Since, corresponding entries of equal matrices are equal, so

$\\\\ \Rightarrow a+b=6 \quad \ldots(i) \\\\ \Rightarrow-2 a+4 b=0 \quad \ldots(i i)\\\\ \Rightarrow c+d=0 \quad \ldots(iii)\\\\ \Rightarrow-2 c+4 d=6 \quad \ldots(iv) \\$

Multiply equation i by 4 and subtract equation ii from i

$\\\\ 4 a+4 b=24\\\\ -2 a+4 b=0\\\\ 6 a=24\\\\ a=\frac{24}{6}\\\\ \Rightarrow a=4$

Put a=4 in equation (i)

$\\\\ \Rightarrow a+b=6\\\\ \Rightarrow 4+b=6\\\\ \Rightarrow b=6-4=2\\\\ \Rightarrow b=2$

Multiply equation iii by 2 and add equation iii and iv

$\\\\ 2 c+2 d=0\\\\ -2 c+4 d=6\\\\ 6 d=6\\\\ \Rightarrow d=1$

Put d=1 in equation iii

$\\\\ \Rightarrow c+d=0\\ \Rightarrow c=-1$

Hence

$A=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]$