# Example 23 - Chapter 3 Class 12 Matrices

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 23 By using elementary operations, find the inverse of the matrix A = [■8(1&2@2&−1)] Given A = [■8(1&2@2&−1)] We know that A = IA [■8(1&2@2&−1)] = [■8(1&0@0&1)] A R2 → R2 – 2R1 [■8(1&2@𝟐−𝟐(𝟏)&−1−2(2))] = [■8(1&0@0−2(1)&1−2(0))] A [■8(1&2@𝟐−𝟐&−1−4)] = [■8(1&0@0−2&1−0)] A [■8(1&2@𝟎&−5)] = [■8(1&0@−2&1)] A R2 → (−1)/5 R2 [■8(1&2@0((−1)/5)&−𝟓((−𝟏)/𝟓) )] = [■8(1&0@−2((−1)/5)&1((−1)/5) )]A [■8(1&2@0&𝟏)] = [■8(1&0@2/5&(−1)/5)]A R1 → R1 – 2R2 [■8(1−2(0)&𝟐−𝟐(𝟏)@0&1)] = [■8(1−2(2/5)&0−2((−1)/5)@2/5&(−1)/5)] A [■8(1 −0&𝟐 −𝟐@0&1)] = [■8(1−4/5&2/5@2/5&(−1)/5)] A [■8(1&𝟎@0&1)] = [■8(1/5&2/5@2/5&(−1)/5)] A I = [■8(1/5&2/5@2/5&(−1)/5)] A This is similar to I = A-1 A Hence A-1 = [■8(1/5&2/5@2/5&(−1)/5)]

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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.