![Example 25 - Find P-1, given P = [10 -2 -5 1] - Matrices](https://d1avenlh0i1xmr.cloudfront.net/f11243c6-a3ad-4377-bb1d-d1314dc9b245/slide72.jpg)
Example 25 - Chapter 3 Class 12 Matrices
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 23 Deleted for CBSE Board 2021 Exams only
Example 24 Important Deleted for CBSE Board 2021 Exams only
Example 25 Important Deleted for CBSE Board 2021 Exams only You are here
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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.