1. Chapter 3 Class 12 Matrices
2. Serial order wise

Transcript

Ex 3.3, 5 For the matrices A and B, verify that (AB)′ = B’A’ where (i) A = [■8(1@− 4@3)] , B = [-1 2 1] Taking L.H.S (AB)’ Finding AB first AB =[■8(1@−4@3)]_(3×1) 〖"[−1 2 1] " 〗_(1×3) = [■8(1×(−1) &1×2&1×1@−4×(−1)&−4×2&−4×1@3×(−1)&3×2&3×1)]_(3×3) = [■8(−1&2&1@4&−8&−4@−3&6&3)] AB = [■8(−1&2&1@4&−8&−4@−3&6&3)] Now, (AB)’= [■8(−1&4&−3@2&−8&6@1&−4&3)] Taking R.H.S B’ A’ Finding B’ , A’ Given B = [− 1 2 1] B’ = 〖"[" − "1 2 1]" 〗^′= [■8(−1@2@1)] Given A = [■8(1@−4@1)] A’ = [1 − 4 1] Now, B’ A’ = [■8(−1@2@1)]_(3×1) 〖"[1 − 4 3]" 〗_(1×3) = [■8(−1×1&−1×(−4)&−1×3@2×1&2×(−4)&2×3@1×1&1×(−4)&1×3)]_(3 × 3) = [■8(−1&4&−3@2&−8&6@1&−4&3)] = L.H.S Hence L.H.S = R.H.S Hence proved 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] Taking 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)] Taking 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

Chapter 3 Class 12 Matrices
Serial order wise