# Ex 3.3, 5 (ii) - Chapter 3 Class 12 Matrices

Last updated at April 16, 2024 by Teachoo

Ex 3.3

Ex 3.3, 1

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)

Chapter 3 Class 12 Matrices

Serial order wise

Last updated at April 16, 2024 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