Question 3 - Inverse of matrix using elementary transformation - Chapter 3 Class 12 Matrices

Last updated at Dec. 16, 2024 by

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

Example 25 - Chapter 3 Class 12 Matrices - Part 2

Go Ad-free

Transcript

Question 3 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.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo