#### Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 13

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: $A=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right], B=\left[\begin{array}{ccc}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]$

Consider,

$A B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right] \\\\ \\ A B=\left[\begin{array}{ccc}0\left(a^{2}\right)+c(a b)+(-b)(a c) & (0)(a b)+(c)\left(b^{2}\right)+(-b)(b c) & (0)(a c)+c(b c)+(-b)\left(c^{2}\right) \\ (-c)\left(a^{2}\right)+(0)(a b)+a(a c) & (-c)(a b)+(0)\left(b^{2}\right)+a(b c) & (-c)(a c)+0(b c)+a\left(c^{2}\right) \\ b\left(a^{2}\right)+(-a)(a b)+0(a c) & (b)(a b)+(-a)\left(b^{2}\right)+0(b c) & (b)(a c)+(-a)(b c)+o\left(c^{2}\right)\end{array}\right]\) \\\\ \\A B=\left[\begin{array}{ccc}0+a b c-a b c & 0+b^{2} c-b^{2} c & 0+b c^{2}-b c^{2} \\ -a^{2} c+0+a^{2} c & -a b c+0+a b c & -a c^{2}+0+a c^{2} \\ a^{2} b-a^{2} b+0 & a b^{2}-a b^{2}+0 & a b c-a b c+0\end{array}\right]$

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

Again consider,

$B A=\left[\begin{array}{ccc}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}\left(a^{2}\right)(0)+(a b)(-c)+(a c)(b) & a^{2}(c)+a b(0)+a c(-a) & a^{2}(-b)+a b(a)+a c(0) \\ a b(0)+b^{2}(-c)+b c(b) & a b(c)+b^{2}(0)+b c(-a) & a b(-b)+b^{2}(a)+b c(0) \\ a c(0)+b c(-c)+c^{2}(b) & a c(c)+b c(0)+c^{2}(-a) & a c(-b)+b c(a)+c^{2}(0)\end{array}\right] \\\\\\ B A=\left[\begin{array}{ccc}0-a b c+a b c & a^{2} c+0-a^{2} c & -a^{2} b+a^{2} b+0 \\ 0-b^{2} c+b^{2} c & a b c+0-a b c & -a b^{2}+a b^{2}+0 \\ 0-b c^{2}+b c^{2} & a c^{2}+0-a c^{2} & -a b c+a b c+0\end{array}\right] \\\\\\ B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \quad \ldots(i i)$

From equation (i) & (ii)

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

Hence, proved