Last updated at Dec. 16, 2024 by Teachoo
In transpose of a matrix,
For matrix
It’s transpose is
We denote it by A’
Similarly for
Let’s look at some properties of transpose
Let’s try to prove them one by one
Let
(A’)’ = A
Therefore,
(A’)’ = A
Therefore,
(4A)’ = 4A’
Therefore,
(AB)’ = B’ A’
For matrix A = [■8(3&2@1&4)] It’s transpose is A’ = [■8(3&1@2&4)] B = [■8(3&2@1&4@5&3)] B’ = [■8(3&1&5@2&4&3)] A = [■8(3&1@2&4)] B = [■8(−8&2@−4&0)] A = [■8(3&2@1&4)] A’ = [■8(3&1@2&4)] (A’)’ = [■8(3&1@2&4)]^′ = [■8(3&2@1&4)] = A A = [■8(3&2@1&4)] Let k = 4 (4A)’ 4A = 4 [■8(3&2@1&4)] = [■8(4×3&4×2@4×1&4×4)] = [■8(12&8@4&16)] 4A’ A = [■8(3&2@1&4)] A’ = [■8(3&1@2&4)] (4A)’ = [■8(12&8@4&16)]^′ = [■8(12&4@8&16)] 4A’ = 4[■8(3&1@2&4)] = [■8(4×3&4×1@4×2&4×4)] = [■8(12&4@8&16)] Let A = [■8(3&1@2&4)], B = [■8(−8&2@−4&0)] (A + B)’ A + B = [■8(3&2@1&4)] + [■8(−8&2@−4&0)] = [■8(3+(−8)&2+2@1+(−4)&4+0)] = [■8(−5&4@−3&4)] A’ + B’ A’ = [■8(3&1@2&4)] B’ = [■8(−8&2@−4&0)] (A + B)’ = [■8(−5&−3@4&4)] A’ + B’ = [■8(3&1@2&4)] + [■8(−8&−4@2&0)] = [■8(3+(−8)&1+(−4)@2+2&4+0)] = [■8(−5&−3@4&4)] Let A = [■8(3&1@2&4)], B = [■8(−8&2@−4&0)] (AB)’ AB = [■8(3&1@2&4)] [■8(−8&2@−4&0)] = [■8(3×(−8)+2×(−4)&3×2+2×0@1×(−8)+4×(−4)&1×2+4×8)] = [■8(−24−8&6+0@−8−16&2+0)] = [■8(−32&6@−24&2)] (AB)’ = [■8(−32&−24@6&2)] B’A’ A = [■8(3&1@2&4)] A’ = [■8(3&1@2&4)] B = [■8(−8&2@−4&0)] B’ = [■8(−8&−4@2&0)] B’A’ =[■8(−8&−4@2&0)][■8(3&1@2&4)] = [■8((−8)×3+(−4)×2&(−8)×1+(−4)×4@2×3+0×2&2×1+0×4)] = [■8(−24−8&−8−16@6+0&2+0)] = [■8(−32&−24@6&2)]
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo