# Example 20 - Chapter 3 Class 12 Matrices

Last updated at April 16, 2024 by Teachoo

Examples

Example 1

Example 2

Example 3 Important

Example 4

Example 5 Important

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20 Important You are here

Example 21

Example 22 Important

Example 23 Important

Example 24 Important

Example 25

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Chapter 3 Class 12 Matrices

Serial order wise

Last updated at April 16, 2024 by Teachoo

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’, Solving 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)] Solving 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 Since 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. Solving 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)] Solving 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 Since L.H.S = R.H.S Hence Proved.