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  • Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4 point 3 question 16 sub question (ii)

Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 16 sub question (ii)  

Answers (1)

Answer:

Hence proved (AB)C = A(BC)

Hint: Associating property of multiplication is (AB)C = A(BC)

Given:

A=\left[\begin{array}{lll} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right], B=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right], C=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]

Consider,

(A B) C=\left(\left[\begin{array}{ccc} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right]\right)\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right] \\

=\left[\begin{array}{ccc} 4(1)+2(0)+3(2) & 4(-1)+2(1)+3(-1) & 4(1)+2(2)+3(1) \\ 1(1)+1(0)+2(2) & 1(-1)+1(1)+2(-1) & 1(1)+1(2)+2(1) \\ 3(1)+0+1(2) & 3(-1)+0(1)+1(-1) & 3(1)+0(2)+1(1) \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]

\\\\ =\left[\begin{array}{ccc} 4+0+6 & -4+2-3 & 4+4+3 \\ 1+0+4 & -1+1-2 & 1+2+2 \\ 3+0+2 & -3+0-1 & 3+0+1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right] \\

=\left[\begin{array}{ccc} 10 & -5 & 11 \\ 5 & -2 & 5 \\ 5 & -4 & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right] \\

= {\left[\begin{array}{ccc} 10(1)+(-5)(3)+11(0) & 10(2)+(-5)(0)+11(0) & 10(-1)+(-5)(1)+11(1) \\ 5(1)+(-2)(3)+5(0) & 5(2)+(-2)(0)+5(0) & 5(-1)+(-2)(1)+5(1) \\ 5(1)+(-4)(3)+4(0) & 5(2)+(-4)(0)+4(0) & 5(-1)+(-4)(1)+4(1) \end{array}\right]} \\\\\\ =\left[\begin{array}{ccc} 10-15+0 & 20+0+0 & -10-5+11 \\ 5-6+0 & 10+0+0 & -5-2+5 \\ 5-12+0 & 10+0+0 & -5-4+4 \end{array}\right]

(A B) C=\left[\begin{array}{ccc} -5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5 \end{array}\right]---- (1)

Now consider RHS

A(B C) =\left[\begin{array}{lll} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left(\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]\right) \\\\\\=\left[\begin{array}{lll} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1-3+0 & 2+0+0 & -1-1+1 \\ 0+3+0 & 0+0+0 & 0+1+2 \\ 2-3+0 & 4-0+0 & -2-1+1 \end{array}\right] \\ \\\\=\left[\begin{array}{ccc} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} -2 & 2 & -1 \\ 3 & 0 & 3 \\ -1 & 4 & -2 \end{array}\right] \\

=\left[\begin{array}{lll} 4(-2)+2(3)+3(-1) & 4(2)+2(0)+3(4) & 4(-1)+2(3)+3(-2) \\ 1(-2)+1(3)+2(-1) & 1(2)+1(0)+2(4) & 1(-1)+1(3)+2(-2) \\ 3(-2)+0(3)+1(-1) & 3(2)+0(0)+1(4) & 3(-1)+0(3)+1(-2) \end{array}\right] \\\\\\ =\left[\begin{array}{lll} -8+6-3 & 8+0+12 & -4+6-6 \\ -2+3-2 & 2+0+8 & -1+3-4 \\ -6+0-1 & 6+0+4 & -3+0-2 \end{array}\right] \\\\\\ A(B C) =\left[\begin{array}{lll} -5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5 \end{array}\right] ... (ii)

From equation (i) and (ii), it is clear that(AB)C = A(BC)

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