# Example 23 - Chapter 3 Class 12 Matrices (Term 1)

Last updated at Dec. 20, 2019 by Teachoo

Last updated at Dec. 20, 2019 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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Example 23 Deleted for CBSE Board 2022 Exams You are here

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Chapter 3 Class 12 Matrices (Term 1)

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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 10 years. He provides courses for Maths and Science at Teachoo.