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Ex 3.4, 15 - Find inverse [2 -3 3 3 2 3 3 -2 2] - Chapter 3 NCERT - Inverse of matrix using elementary transformation

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise
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Ex3.4, 15 Find the inverse of each of the matrices, if it exists. [■8(2&−3&3@2&2&3@3&−2&2)] Let A =[■8(2&−3&3@2&2&3@3&−2&2)] We know that A = IA [■8(2&−3&3@2&2&3@3&−2&2)]= [■8(1&0&0@0&1&0@0&0&1)] A R1 →R_1/2 , [■8(𝟐/𝟐&(−3)/2&3/2@2&2&3@3&−2&2)]= [■8(1/2&0/2&0/2@0&1&0@0&0&1)] A [■8(𝟏&(−3)/2&3/2@2&2&3@3&−2&2)]= [■8(1/2&0&0@0&1&0@0&0&1)] A R2 → R2 – 2R1 [■8(1&(−3)/2&3/2@𝟐−𝟐(𝟏)&2−2((−3)/2)&3−2(3/2)@3&−2&2)] = [■8(1/2&0&0@0−2(1/2)&1−2(0)&0−2(0)@0&0&1)] A [■8(1&(−3)/2&3/2@𝟎&5&0@3&−2&2)] = [■8(1/2&0&0@−1&1&0@0&0&1)] A R3 → R3 – 3R1 [■8(1&(−3)/2&3/2@0&5&0@𝟑−𝟑(𝟏)&−2−3((−3)/2)&2−3(3/2) )]= [■8(1/2&0&0@−1&1&0@0−3(1/2)&0−3(0)&1−3(0))] A [■8(1&(−3)/2&3/2@0&5&0@𝟎&5/2&(−5)/2)]= [■8(1/2&0&0@−1&1&0@(−3)/2&0&1)] A R2 →R_2/5 , [■8(1&(−3)/2&3/2@0&𝟓/𝟓&0@0&5/2&(−5)/2)] = [■8(1/2&0&0@(−1)/5&1/5&0@(−3)/2&0&1)] A [■8(1&(−3)/2&3/2@0&𝟏&0@0&5/2&(−5)/2)] = [■8(1/2&0&0@(−1)/5&1/5&0@(−3)/2&0&1)] A R1 → R1 + 3/2 R2 [■8(1+3/2(0)&(−𝟑)/𝟐+𝟏(𝟑/𝟐)&3/2+3/2(0)@0&1&0@0&5/2&(−5)/2)]= [■8(1/2+3/2 ((−1)/5)&0+3/2 (1/5)&0+3/2(0)@(−1)/5&1/5&0@(−3)/2&0&1)] A [■8(1&𝟎&3/2@0&1&0@0&5/2&(−5)/2)]= [■8(2/10&3/10&0@(−1)/5&1/5&0@(−3)/2&0&1)] A R3 → R3 – 5/2 R2 [■8(1&0&3/2@0&1&0@0−5/2(0)&𝟓/𝟐−𝟓/𝟐(𝟏)&(−5)/2−(−5)/2(0))]= [■8(1/5&3/10&0@(−1)/5&1/5&0@(−3)/2−5/2 (−1/2)&0 (−5)/2 (1/5)&1 (−5)/2(0))] A [■8(1&0&3/2@0&1&0@0&𝟎&(−5)/2)] = [■8(1/5&3/10&0@(−1)/5&1/5&0@1&(−1)/2&1)] A R3 → (−2)/5 R3 [■8(1&0&3/2@0&1&0@0((−2)/5)&0(−2/5)&(−𝟓)/𝟐 ((−𝟐)/𝟓) )]= [■8(1/5&3/10&0@(−1)/5&1/5&0@1((−2)/5)&(−1)/2 ((−2)/5)&1((−2)/5) )] A [■8(1&0&3/2@0&1&0@0&0&𝟏)] = [■8(1/5&3/10&0@(−1)/5&1/5&0@(−2)/5&1/5&(−2)/5)] A R1 → R1 −3/2 R3 [■8(1−3/2(0)&0−3/2(0)&𝟑/𝟐−𝟑/𝟐(𝟏)@0&1&0@0&0&1)]= [■8(1/5−3/2 (2/5)&3/10−3/2 (1/5)&0−3/2 ((−2)/5)@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A [■8(1&0&𝟎@0&1&0@0&0&1)] = [■8((−2)/5&0&3/5@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A I = [■8((−2)/5&0&3/5@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A This is similar to I = A-1A Thus, A-1 =[■8((−2)/5&0&3/5@(−1)/5&1/5&0@2/5&1/5&(−2)/5)]

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