# Example 20 - Chapter 3 Class 12 Matrices

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 20 If A = [■8(3&√3&2@4&2&0)] and B = [■8(2&−1&2@1&2&4)] Verify that (i) (A’)’ = A, A = [■8(3&√3&2@4&2&0)] A’ = [■8(3&√3&2@4&2&0)]^′= [■8(3&4@√3&2@2&0)] (A’)’ = [■8(3&4@√3&2@2&0)]^′= [■8(3&√3&2@4&2&0)] = A Thus (A’)’ = A Example 20 If A = [■8(3&√3&2@4&2&0)] and B = [■8(2&−1&2@1&2&4)] Verify that (ii) (A + B)’ = A’ + B’, Taking L.H.S First finding (A + B) (A + B) = [■8(3&√3&2@4&2&0)] + [■8(2&−1&2@1&2&4)] = [■8(3+2 &√3+(−1)&2+2@4+1&2+2&0+4)] = [■8(5&√3−1&4@5&4&4)] Thus, (A + B)’ = [■8(5&5@√3−1&4@4&4)] Taking R.H.S A’ + B’ Finding A’ A = [■8(3&√3&2@4&2&0)] A’ = [■8(3&4@√3&2@2&0)] Also, B = [■8(2&−1&2@1&2&4)] B‘ = [■8(2&1@−1&2@2&4)]Now, A’ + B’ =[■8(3&4@√3&2@2&0)] +[■8(2&1@−1&2@2&4)] = [■8(3+2&4+1@√3+(−1)&2+2@2+0&0+4)] = [■8(5&5@√3−1&4@4&4)] = L.H.S Hence, L.H.S = R.H.S Hence Proved Example 20, If A = [■8(3&√3&2@4&2&0)] and B = [■8(2&−1&2@1&2&4)] .Verify that (iii) (kB)’ = kB’, where k is any constant. Taking L.H.S (kB)’ Finding kB first kB = k [■8(2&−1&2@1&2&4)] = [■8(2k&−k&2k@k&2k&4k)] (kB)’ = [■8(2k&k@−𝑘&2𝑘@2k&4k)] Taking R.H.S kB’ Finding B’ first B = [■8(2&−1&2@1&2&4)] B’ = [■8(2&1@−1&2@2&4)] kB’ = k[■8(2&1@−1&2@2&4)] = [■8(2k&k@−k&2k@2k&4k)] = L.H.S Hence, L.H.S = R.H.S Hence Proved.

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18 Important

Example 19 Important

Example 20 You are here

Example 21

Example 22 Important

Example 23

Example 24

Example 25

Example 26

Example 27 Important

Example 28 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.