Ex 3.4

Chapter 3 Class 12 Matrices
Serial order wise

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### Transcript

Ex 3.4, 16 Find the inverse of each of the matrices, if it exists. [■8(1&3&−[email protected]−3&0&−[email protected]&5&0)] Let A = [■8(1&3&−[email protected]−3&0&−[email protected]&5&0)] A = IA [■8(1&3&−[email protected]−3&0&−[email protected]&5&0)] = [■8(1&0&[email protected]&1&[email protected]&0&1)] A R2 → R2 + 3R1 [■8(1&3&−[email protected]−𝟑+𝟑(𝟏)&0+3(3)&−5+3(−2)@2&5&0)] = [■8(1&0&[email protected]+3(1)&1+3(0)&0+3(0)@0&0&1)] A [■8(1&3&−[email protected]𝟎&9&−[email protected]&5&0)] = [■8(1&0&[email protected]&1&[email protected]&0&1)] A R3 → R3 − 2R1 [■8(1&3&−[email protected]&9&−[email protected]𝟐−𝟐(𝟏)&5−2(3)&0−2(−2))] = [■8(1&0&[email protected]&1&[email protected]−2(1)&0−2(0)&1−2(0))] A [■8(1&3&−[email protected]&9&−[email protected]𝟎&−1&4)] = [■8(1&0&[email protected]&1&[email protected]−2&0&1)] A R2 → 1/9 R2 [■8(1&3&−[email protected]/9&𝟗/𝟗&(−11)/[email protected]&−1&4)]= [■8(1&0&[email protected]/9&1/9&0/[email protected]−2&0&1)]A [■8(1&3&−[email protected]&𝟏&(−11)/[email protected]&−1&4)] = [■8(1&0&[email protected]/3&1/9&[email protected]−2&0&1)] A R3 → R3 + R2 [■8(1&3&−[email protected]&1&(−11)/[email protected]+0&−𝟏+𝟏&4+((−11)/9) )] = [■8(1&0&[email protected]/3&1/9&[email protected]−2+1/3&0+1/9&1+0)] A [■8(1&3&−[email protected]&1&(−11)/[email protected]&𝟎&25/9)] = [■8(1&0&[email protected]/3&1/9&[email protected](−5)/3&1/9&1)] A R1 → R1 − 3R2 [■8(1−3(0)&𝟑−𝟑(𝟏)&−2−3((−11)/9)@0&1&(−11)/[email protected]&0&25/3)] =[■8(1−3(1/3)&0−3(1/9)&0−3(0)@1/3&1/9&[email protected](−5)/3&1/9&1)] A ` [■8(1&𝟎&5/[email protected]&1&(−11)/[email protected]&0&25/9)] = [■8(0&(−1)/3&[email protected]/3&1/9&[email protected](−5)/3&1/9&1)] A R3 → 9/25 R3 [■8(1&0&5/[email protected]&1&(−11)/[email protected]&0&𝟏)] = [■8(0&(−1)/3&[email protected]/3&1/9&[email protected](−3)/5&1/25&9/25)] A R2 → R2 + 11/9 R3 [■8(1&0&5/[email protected]+11/9 (0)&1+11/9(0)&(−𝟏𝟏)/𝟗+𝟏𝟏/𝟗(𝟏)@0&0&1)] =[■8(0&(−1)/3&[email protected]/3+11/9 ((−3)/5)&1/9+11/9 (1/25)&0+11/9 (9/25)@(−3)/5&1/25&9/25)] A [■8(1&0&5/[email protected]&1&𝟎@0&0&1)] = [■8(0&(−1)/3&[email protected](−2)/5&4/25&11/[email protected](−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&[email protected]&0&1)] =[■8(0−5/3 ((−3)/5)&(−1)/3− 5/3 (1/25)&0−5/3 (9/25)@(−2)/5&4/25&11/[email protected](−3)/5&1/25&9/25)] A [■8(1&0&𝟎@0&1&[email protected]&0&1)] = [■8(5/5&(−6)/15 &(−3)/[email protected](−2)/5&4/25&11/[email protected](−3)/5&1/25&9/25)] A I = [■8(1&(−2)/5&(−3)/[email protected](−2)/5&4/25&11/[email protected](−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&𝟏𝟏/𝟐𝟓@(−𝟑)/𝟓&𝟏/𝟐𝟓&𝟗/𝟐𝟓)]