![Example 25 - Find P-1, given P = [10 -2 -5 1] - Matrices](https://d1avenlh0i1xmr.cloudfront.net/f11243c6-a3ad-4377-bb1d-d1314dc9b245/slide72.jpg)
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Example 25 - Chapter 3 Class 12 Matrices (Term 1)
Last updated at Jan. 17, 2020 by Teachoo
Transcript
Example 25 Find P -1, if it exists, given P = [■8(10&−2@−5&1)] Given P = [■8(10&−2@−5&1)] We know that P = I P [■8(10&−2@−5&1)] = [■8(1&0@0&1)] P R1 →1/10 R1 [■8(𝟏𝟎/𝟏𝟎&(−2)/10@−5&1)]" = " [■8(1/10&0/10@0&1)]" P" [■8(𝟏&(−1)/5@−5&1)] = [■8(1/10&0@0&1)] P R2 →"R2" + 5R1 [■8(1&(−1)/5@−𝟓+𝟓(𝟏)&1+5((−1)/5) )]" = " [■8(1/10&0@0+5(1/10)&1+5(0))]" P" [■8(1&(−1)/5@−𝟓+𝟓&1−1)]" = " [■8(1/10&0@0+(1/2)&1+0)]" P" [■8(1&(−1)/5@𝟎&0)] = [■8(1/10&0@1/2&1)] P Since we have all zeros in the second row of the left hand side matrix of the above equation. Therefore, P–1 does not exist.
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Example 25 Important Deleted for CBSE Board 2022 Exams You are here
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Chapter 3 Class 12 Matrices (Term 1)
Serial order wise
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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.