Transpose of a matrix
Last updated at April 16, 2024 by Teachoo
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] Solving 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)] Solving 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