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Ex 3.4, 6
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex3.4, 6 Find the inverse of each of the matrices, if it exists.[■8(2&5@1&3)] Let A = [■8(2&5@1&3)] We know that A = IA [■8(2&5@1&3)] = [■8(1&0@0" " &1)] A R1→ R1 – R2 [■8(𝟐−𝟏&5−3@1&3)] = [■8(1−0&0−1@0" " &1)] A [■8(𝟏&2@1&3)] = [■8(1&−1@0" " &1)] A R2→ R2 – R1 [■8(1&2@𝟏−𝟏&3−2)] = [■8(1&−1@0" " −1&1−(−1))] A [■8(1&2@𝟎&1)] = [■8(1&−1@− 1&2)] A R1 → R1 – 2R2 [■8(1−2(0)&𝟐−𝟐(𝟏)@0&1)] = [■8(1−2(−1)&−1−2(2)@− 1&2)] A [■8(1&𝟎@0&1)] = [■8(3&−5@−1&2)] A I = [■8(3&−5@−1&2)] A This is similar to I = A-1A Thus, A-1 = [■8(3&−5@−1&2" " )]
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
