#### Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question  48 sub question (vi)

$A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0 \\ -5 & 4\end{array}\right]$

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:

$A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]$

The matrix given on the RHS of the equation is a $3 \times 3$ matrix and the one given on the LHS of the equation is $2 \times 3$ matrix. So,  A has to be $3 \times 2$ matrix.

Now, let

$\\\\A=\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right] \\\\$

$\\\\\Rightarrow\left[\begin{array}{lll} a(1)+b(4) & a(2)+b(5) & a(3)+b(6) \\ c(1)+d(4) & c(2)+5(d) & c(3)+d(6) \\ e(1)+4(f) & e(2)+5(f) & 3(e)+6(f) \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} a+4 b & 2 a+5 b & 3 a+6 b \\ c+4 d & 2 c+5 d & 3 c+6 d \\ e+4 f & 2 e+5 f & 3 e+6 f \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{array}\right]\\$

Equating the corresponding elements of the two matrices, we have

$a + 4b = -7$           ...(i)

$2a +5b = -8$       ...(ii)

Now multiply equation i by 2 and subtract equation ii from i

$\begin{array}{l} 2 a+8 b=-14\\ 2 a+5 b=-8\\ 3 b=-6\\ \Rightarrow b=-2\\ \end{array}$

Put value of b=-2 in equation ii we get

$2a + 5 (-2) = -8$

$\\\\ \Rightarrow 2 a-10=-8\\ \Rightarrow 2 a=-8+10\\ \Rightarrow 2 a=2\\ \Rightarrow a=1\\\$

$\\\\ c+4 d=2 ...(iii) \\\\ \Rightarrow 2 c+5 d=4 ... (iv)$

Multiply equation iii by 2 and subtract equation iv from iii

$\\\\ 2 c+8 d=4\\\\ \underline{2 c+5 d=4}\\\\$

$3 d=0 \Rightarrow d=0$

Now, put d=0 in equation iv, we get

$\\\\ \Rightarrow 2 c+5(0)=4\\\\ \Rightarrow 2 c=4\\\\ \Rightarrow c=2\\\\ \Rightarrow e+4 f=11 \cdots (v)\\\\ \Rightarrow 2 e+5 f=10 \cdots(vi) \\\\$

Multiply equation v by 2 and subtract equation vi from v

$\begin{array}{l} 2 e+8 f=22\\\\ \underline{2 e+5 f=10}\\\\ \Rightarrow 3 f=12\\ \\\Rightarrow f=4\\\\ \text { Put } f=4 \text { in equation }(v i)\\\\ \Rightarrow 2 e+5(4)=10\\\\ \Rightarrow 2 e+20=10\\\\ \Rightarrow 2 e=10-20\\\\ \Rightarrow 2 e=-10\\\\ \Rightarrow e=-5\\ \end{array}$

Thus,

$a=1, b=-2, c=2, d=0, e-5 \text { and } f=4$

Hence,

$A=\left[\begin{array}{cc} 1 & -2 \\ 2 & 0 \\ -5 & 4 \end{array}\right]$