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Ex 3.4, 16 - Find inverse of matrix [1 3 -2 -3 0 5 - Inverse of matrix using elementary transformation

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise
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Ex3.4, 16 Find the inverse of each of the matrices, if it exists. [■8(1&3&−2@−3&0&−5@2&5&0)] Let A = [■8(1&3&−2@−3&0&−5@2&5&0)] A = IA [■8(1&3&−2@−3&0&−5@2&5&0)] = [■8(1&0&0@0&1&0@0&0&1)] A R2 → R2 + 3R1 [■8(1&3&−2@−𝟑+𝟑(𝟏)&0+3(3)&−5+3(−2)@2&5&0)] = [■8(1&0&0@0+3(1)&1+3(0)&0+3(0)@0&0&1)] A [■8(1&3&−2@𝟎&9&−11@2&5&0)] = [■8(1&0&0@3&1&0@0&0&1)] A R3 → R3 − 2R1 [■8(1&3&−2@0&9&−11@𝟐−𝟐(𝟏)&5−2(3)&0−2(−2))] = [■8(1&0&0@3&1&0@0−2(1)&0−2(0)&1−2(0))] A [■8(1&3&−2@0&9&−11@𝟎&−1&4)] = [■8(1&0&0@3&1&0@−2&0&1)] A R2 → 1/9 R2 [■8(1&3&−2@0/3&𝟗/𝟗&(−11)/9@0&−1&4)]= [■8(1&0&0@3/9&1/9&0/9@−2&0&1)]A [■8(1&3&−2@0&𝟏&(−11)/9@0&−1&4)] = [■8(1&0&0@1/3&1/9&0@−2&0&1)] A R3 → R3 − R2 [■8(1&3&−2@0&1&(−11)/9@0+0&−𝟏+𝟏&4+((−11)/9) )] = [■8(1&0&0@1/3&1/9&0@−2+1/3&0+1/9&1+0)] A [■8(1&3&−2@0&1&(−11)/9@0&𝟎&25/9)] = [■8(1&0&0@1/3&1/9&0@(−5)/3&1/9&1)] A R1 → R1 − 3R2 [■8(1−3(0)&𝟑−𝟑(𝟏)&−2−3(11/3)@0&1&(−11)/9@0&0&25/3)] =[■8(1−3(1/3)&0−3(1/9)&0−3(0)@1/3&1/9&0@(−5)/3&1/9&1)] A [■8(1&𝟎&5/3@0&1&(−11)/9@0&0&25/9)] = [■8(0&(−1)/3&0@1/3&1/9&0@(−5)/3&1/9&1)] A R3 → 9/25 R3 [■8(1&0&5/3@0&1&(−11)/9@0&0&𝟏)] = [■8(0&(−1)/3&0@1/3&1/9&0@(−3)/5&1/25&9/25)] A R2 → R2 + 11/9 R3 [■8(1&0&5/3@0+11/9 (0)&1+11/9(0)&(−𝟏𝟏)/𝟗+𝟏𝟏/𝟗(𝟏)@0&0&1)] =[■8(0&(−1)/3&0@1/3+11/9 ((−3)/5)&1/9+11/9 (1/25)&0+11/9 (9/25)@(−5)/3&1/25&9/25)] A [■8(1&0&5/3@0&1&𝟎@0&0&1)] = [■8(0&(−1)/3&0@−(−2)/5&1/9+1/(9×25)&1/25@(−3)/5&1/25&9/25)] A R1 → R1 – 5/3 R3 [■8(1+ 5/3 (0)&0+ 5/3 (0)&(−𝟓)/𝟑+ 𝟓/𝟑 (𝟏)@0&1&0@0&0&1)] =[■8(0−5/3 ((−3)/5)&(−1)/3− 5/3 ((−3)/5)&0−5/3 (9/25)@2/5&4/25&1/25@(−3)/5&1/25&9/25)] A [■8(1&0&𝟎@0&1&0@0&0&1)] = [■8((−5)/5&(−1)/3− 1/( 5×3) &3/5@2/5&(−4)/25&11/25@(−3)/5&1/25&9/25)] A I = [■8(−1&(−2)/5&3/5@2/5&(−4)/25&1/25@(−3)/5&1/25&9/25)] A This is similar to I = A-1 A Hence A-1 = [■8(−1&(−2)/5&3/5@2/5&(−4)/25&1/25@(−3)/5&1/25&9/25)]

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