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  • Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question12 maths textbook solution

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

Answers (1)

Answer:

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

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