For two matrices 𝐴=[(3 −6 −1 2 −5 −1 −2 4 1)] and 𝐵=[(1 −2 −1 0 −1

Question 32 For two matrices 𝐴=[■(3&−6&−1@2&−5&−1@−2&4&1)] and 𝐵=[■(1&−2&−1@0&−1&−1@2&0&3)], find the product 𝐴𝐵 and hence solve the system of equations: 3𝑥−6𝑦−𝑧=3 2𝑥−5𝑦−𝑧+2=0 −2𝑥+4𝑦+𝑧=5Finding the product AB = [■(3&−6&−1@2&−5&−1@−2&4&1)] [■(1&−2&−1@0&−1&−1@2&0&3)] =[■8(3(1)+(⤶7−6)(0)+(−1)(2)&3(−2)+(−6)(−1)+(−1)(0)&3(−1)+(−6)(−1)+(−1)(3)@2(1)+(−5)(0)+(−1)(2)&2(−2)+(−5)(−1)+(−1)(0)&2(−1)+(−5)(−1)+(−1)(3)@(−2)(1)+4(0)+1(2)&(−2)(−2)+4(−1)+1(0)&(−2)(−1)+4(−1)+1(3))] = [■8(1@0@0)" " ■8(0@1@0)" " ■8(0@0@1)] Thus, AB = I We know that AA-1 = I So 𝑩 is inverse of A Now, solving the equation Given equations are 3𝑥−6𝑦−𝑧=3 2𝑥−5𝑦−𝑧=−2 −2𝑥+4𝑦+𝑧=5 Writing the equation as AX = D [■(3&−6&−1@2&−5&−1@−2&4&1)][■8(𝑥@𝑦@𝑧)] = [■8(3@−2@5)] Here A =[■(3&−6&−1@2&−5&−1@−2&4&1)], X = [■8(𝑥@𝑦@𝑧)] & D = [■8(3@−2@5)] Now, AX = D X = A-1 D Putting A-1 = 𝑩=[■(1&−2&−1@0&−1&−1@2&0&3)] So, our equation becomes [■8(𝑥@𝑦@𝑧)] =[■(1&−2&−1@0&−1&−1@2&0&3)][■8(3@−2@5)] [■8(𝑥@𝑦@𝑧)] = [■8(1(3)+(⤶7−2)(−2)+(−1) (5)@0(3)+(−1)(−2)+(−1)(5)@2(3)+0(−2)+3(5))] [■8(𝑥@𝑦@𝑧)] = [■8(3+4−5@0+2−5@6+0+15)] [■8(𝑥@𝑦@𝑧)] = [■8(2@−3@21)] Hence x = 2 , y = −3 & z = 21

Question 32 - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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