#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 14 math textbook solution.

Answer: Hence, proved $A B=A$  and $B A=B$

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

Consider,

$A B=\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right] \\\\ A B=\left[\begin{array}{ccc}2(2)+(-3)(-1)+(-5)(1) & 2(-2)+(-3)(3)+(-5)(-2) & 2(-4)+(-3)(4)+(-5)(-3) \\ (-1)(2)+4(-1)+5(1) & (-1)(-2)+4(3)+5(-2) & (-1)(-4)+4(4)+5(-3) \\ (1)(2)+(-3)(-1)+(-4)(1) & 1(-2)+(-3)(3)+(-4)(-2) & (1)(-4)+(-3)(4)+(-4)(-3)\end{array}\right] \\\\\\ A B=\left[\begin{array}{ccc}4+3-5 & -4-9+10 & -8-12+15 \\ -2-4+5 & 2+12-10 & 4+16-15 \\ 2+3-4 & -2-9+8 & -4-12+12\end{array}\right]$

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

Therefore AB=A

Again consider,

$B A=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]\left[\begin{array}{ccc}2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}2(2)+(-2)(-1)+(-4)(1) & 2(-3)+(-2)(4)+(-4)(-3) & 2(-5)+(-2)(5)+(-4)(-4) \\ -1(2)+3(-1)+4(1) & (-1)(-3)+3(4)+4(-3) & (-1)(-5)+3(5)+4(-4) \\ 1(2)+(-2)(-1)+(-3)(1) & (1)(-3)+(-2)(4)+(-3)(-3) & (1)(-5)+(-2)(5)+(-3)(-4)\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}4+2-4 & -6-8+12 & -10-10+16 \\ -2-3+4 & 3+12-12 & 5+15-16 \\ 2+2-3 & -3-8+9 & -5-10+12\end{array}\right] \\\\ \\\\B A=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]=B \\\\ B A=B$

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