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Example 23 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at March 30, 2023 by Teachoo
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Example 23 You are here
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Chapter 3 Class 12 Matrices
Serial order wise
Transcript
Example 23 By using elementary operations, find the inverse of the matrix A = [■8(1&[email protected]&−1)] Given A = [■8(1&[email protected]&−1)] We know that A = IA [■8(1&[email protected]&−1)] = [■8(1&[email protected]&1)] A R2 → R2 – 2R1 [■8(1&[email protected]𝟐−𝟐(𝟏)&−1−2(2))] = [■8(1&[email protected]−2(1)&1−2(0))] A [■8(1&[email protected]𝟐−𝟐&−1−4)] = [■8(1&[email protected]−2&1−0)] A [■8(1&[email protected]𝟎&−5)] = [■8(1&[email protected]−2&1)] A R2 → (−1)/5 R2 [■8(1&[email protected]((−1)/5)&−𝟓((−𝟏)/𝟓) )] = [■8(1&[email protected]−2((−1)/5)&1((−1)/5) )]A [■8(1&[email protected]&𝟏)] = [■8(1&[email protected]/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/[email protected]/5&(−1)/5)] A [■8(1&𝟎@0&1)] = [■8(1/5&2/[email protected]/5&(−1)/5)] A I = [■8(1/5&2/[email protected]/5&(−1)/5)] A This is similar to I = A-1 A Hence A-1 = [■8(1/5&2/[email protected]/5&(−1)/5)]
