#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (i) maths textbook solution

$A=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 0 & 1\end{array}\right]$

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:

$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] A=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$

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

Since,

$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]_{2 \times 2}\ \ \ \ \ \left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]_{2 \times 3}$

So,A is a matrix of order $2 \times 3$

So, let

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

$\\\\ \Rightarrow\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right] \\\\ \Rightarrow\left[\begin{array}{lll}a+d & b+e & c+f \\ 0+d & 0+e & 0+f\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$

Since, corresponding entries of equal matrices are equal,

So,

$d=1, c=0, f=1$

and

$\\\\a+d=3 \Rightarrow a+1=3 \Rightarrow a=3-1 \Rightarrow a=2 \\\\ b+e=3 \Rightarrow b+0=3 \Rightarrow b=3$

And

$c+f=5 \Rightarrow c+1=5 \Rightarrow c=4$

Hence,

$\\\\A=\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right] \\\\ A=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 0 & 1\end{array}\right]$