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

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

Answers (1)

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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