#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 12 maths textbook solution

Hence, prove $A B=B A=0_{3 \times 3}$

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:

$A=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right], B=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]$

$\text { Prove: } A B=B A=0_{3 \times 3}$

Consider

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

$A B=\left[\begin{array}{ccc}-2-3+5 & 6+9-15 & 10+15-25 \\ 1+4-5 & -3-12+15 & -5-20+25 \\ -1-3+4 & 3+9-12 & 5+15-20\end{array}\right]$

$A B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] . . . . (i)$

Again consider

$B A=\left[\begin{array}{ccc}-1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5\end{array}\right]\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\\\\\\ B A=\left[\begin{array}{ccc}(-1)(2)+(3)(-1)+5(1) & (-1)(-3)+(3)(4)+5(-3) & (-1)(-5)+3(5)+5(-4) \\ (1)(2)+(-3)(-1)+(-5)(1) & (1)(-3)+(-3)(4)+(-5)(-3) & (1)(-5)+(-3)(5)+(-5)(-4) \\ (-1)(2)+(3)(-1)+(5)(1) & (-1)(-3)+(3)(4)+(5)(-3) & (-1)(-5)+(3)(5)+(5)(-4)\end{array}\right] \\\\\\\ B A=\left[\begin{array}{ccc}-2-3+5 & 3+12-15 & 5+15-20 \\ 2+3-5 & -3-12+15 & -5-15+20 \\ -2-3+5 & 3+12-15 & 5+15-20\end{array}\right] \\\\ \\B A=\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \quad \ldots(i i)$

From equation (i) & (ii)

$A B=B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]_{3 \times 3}$

Hence, proved