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Example 25 - Find P-1, given P = [10 -2 -5 1] - Matrices

Example 25 - Chapter 3 Class 12 Matrices - Part 2

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Example 25 Find P -1, if it exists, given P = [■8(10&−[email protected]−5&1)] Given P = [■8(10&−[email protected]−5&1)] We know that P = I P [■8(10&−[email protected]−5&1)] = [■8(1&[email protected]&1)] P R1 →1/10 R1 [■8(𝟏𝟎/𝟏𝟎&(−2)/[email protected]−5&1)]" = " [■8(1/10&0/[email protected]&1)]" P" [■8(𝟏&(−1)/[email protected]−5&1)] = [■8(1/10&[email protected]&1)] P R2 →"R2" + 5R1 [■8(1&(−1)/[email protected]−𝟓+𝟓(𝟏)&1+5((−1)/5) )]" = " [■8(1/10&[email protected]+5(1/10)&1+5(0))]" P" [■8(1&(−1)/[email protected]−𝟓+𝟓&1−1)]" = " [■8(1/10&[email protected]+(1/2)&1+0)]" P" [■8(1&(−1)/[email protected]𝟎&0)] = [■8(1/10&[email protected]/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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.