![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 (Term 1)
Last updated at Jan. 17, 2020 by
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 6
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Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21
Example 22 Important
Example 23 Deleted for CBSE Board 2022 Exams
Example 24 Important Deleted for CBSE Board 2022 Exams
Example 25 Important Deleted for CBSE Board 2022 Exams You are here
Example 26 Important
Example 27 Important
Example 28
Chapter 3 Class 12 Matrices (Term 1)
Serial order wise
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.
