#### Please solve RD Sharma class 12 Chapter Algebra of Matrices  exercise 4.3 question 2 sub question (i) maths textbook solution.

Hence proved $AB\neq BA$

$\begin{bmatrix} 7 &1 \\ 33& 34 \end{bmatrix}\neq \begin{bmatrix} 16 &5 \\ 39& 25 \end{bmatrix}$

Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A=\begin{bmatrix} 5 &-1 \\ 6 &7 \end{bmatrix}$ and $B=\begin{bmatrix} 2 &1 \\ 3 &4 \end{bmatrix}$

First, we multiply AB matrix

$AB=\begin{bmatrix} 5 &-1 \\ 6 &7 \end{bmatrix}\begin{bmatrix} 2 &1 \\ 3& 4 \end{bmatrix}$

$A B=\left[\begin{array}{cc} 5 \times 2+(-1) \times 3 & 5 \times 1+(-1) \times 4 \\ 6 \times 2+7 \times 3 & 6 \times 1+7 \times 4 \end{array}\right]$

$AB=\begin{bmatrix} 10-3 &5-4 \\ 12+21& 6+28 \end{bmatrix}$

$AB=\begin{bmatrix} 7 &1 \\ 33& 34 \end{bmatrix}$                                          ...(i)

Again consider

$BA=\begin{bmatrix} 2 &1 \\ 3& 4 \end{bmatrix}\begin{bmatrix} 5 &-1 \\ 6& 7 \end{bmatrix}$

\begin{aligned} &B A=\left[\begin{array}{ll} 2 \times 5+1 \times 6 & 2 \times(-1)+1 \times 7 \\ 3 \times 5+4 \times 6 & 3 \times(-1)+7 \times 4 \end{array}\right] \\ &B A=\left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right] \end{aligned}

$BA =\begin{bmatrix} 16 &5 \\ 39& 25 \end{bmatrix}$                              ...(ii)

From equation (i) and (ii), it is clear that $AB\neq BA$