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Ex 3.3
Ex 3.3, 2
Ex 3.3, 3
Ex 3.3, 4 Important
Ex 3.3, 5 (i)
Ex 3.3, 5 (ii) You are here
Ex 3.3, 6 (i)
Ex 3.3, 6 (ii) Important
Ex 3.3, 7 (i)
Ex 3.3, 7 (ii) Important
Ex 3.3, 8
Ex 3.3, 9
Ex 3.3, 10 (i) Important
Ex 3.3, 10 (ii)
Ex 3.3, 10 (iii) Important
Ex 3.3, 10 (iv)
Ex 3.3, 11 (MCQ) Important
Ex 3.3, 12 (MCQ)
Last updated at June 8, 2023 by Teachoo
Ex 3.3, 5 For the matrices A and B, verify that (AB)′= B′A′, where (ii) A = [■8(0@1@2)] , B = [1 5 7] Solving L.H.S (AB)’ Finding AB AB = [■8(0@1@2)]_(3 × 1) 〖"[1 5 7]" 〗_(1 × 3) = [■8(0×1&0×5&0×7@1×1&1×5&1×7@2×1&2×5&2×7)]_(3×3) = [■8(𝟎&𝟎&𝟎@𝟏&𝟓&𝟕@𝟐&𝟏𝟎&𝟏𝟒)] Thus, AB = [■8(0&0&0@1&5&7@2&10&14)] So, (AB)’ = [■8(𝟎&𝟏&𝟐@𝟎&𝟓&𝟏𝟎@𝟎&𝟕&𝟏𝟒)] Solving R.H.S (B’ A’) Finding B’ B = [1 5 7] B’ = [■8(𝟏@𝟓@𝟕)] Also, A = [■8(0@1@2)] A’ = [0 1 2] B’ A’= [■8(1@5@7)]_(3×1) 〖"[0 1 2] " 〗_(1×3) = [■8(1×0&1×1&1×2@5×0&5×1&5×2@7×0&7×1&7×2)]_(3 × 3) = [■8(𝟎&𝟏&𝟐@𝟎&𝟓&𝟏𝟎@𝟎&𝟕&𝟏𝟒)] = L.H.S Hence L.H.S = R.H.S Hence proved