Provide solution for RD Sharma maths class12 Chapter Algebra of Matrices exercise 4.3 question 3 sub question (i)

$AB=\begin{bmatrix} -3 & -4 &1 \\ 8& 13 &9 \end{bmatrix}$ and BA does not exist.

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

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

Consider,

$AB=\begin{bmatrix} 1 &-2 \\ 2& 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3 \\ 2& 3 &1 \end{bmatrix}$

\begin{aligned} &A B=\left[\begin{array}{ccc} 1 \times 1+(-2) \times 2 & 1 \times 2+(-2) \times 3 & 1 \times 3+(-2) \times 1 \\ 2 \times 1+3 \times 2 & 2 \times 2+3 \times 3 & 2 \times 3+3 \times 1 \end{array}\right] \\ &A B=\left[\begin{array}{ccc} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{array}\right] \end{aligned}

$AB=\begin{bmatrix} -3 &-4 &1 \\ 8& 13 &9 \end{bmatrix}$

Now consider BA

$BA=\begin{bmatrix} 1 &2 &3 \\ 2& 3 &1 \end{bmatrix}\begin{bmatrix} 1 &-2 \\ 2& 3 \end{bmatrix}$

BA doesn’t exist because the number of columns in B is greater than the rows in A.