Ex 3.4, 15 - Chapter 3 Class 12 Matrices (Term 1)

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Ex 3.4, 15 Find the inverse of each of the matrices, if it exists. [■8(2&−3&[email protected]&2&[email protected]&−2&2)] Let A =[■8(2&−3&[email protected]&2&[email protected]&−2&2)] We know that A = IA [■8(2&−3&[email protected]&2&[email protected]&−2&2)]= [■8(1&0&[email protected]&1&[email protected]&0&1)] A R1 →R_1/2 [■8(𝟐/𝟐&(−3)/2&3/[email protected]&2&[email protected]&−2&2)]= [■8(1/2&0/2&0/[email protected]&1&[email protected]&0&1)] A [■8(𝟏&(−3)/2&3/[email protected]&2&[email protected]&−2&2)]= [■8(1/2&0&[email protected]&1&[email protected]&0&1)] A R2 → R2 – 2R1 [■8(1&(−3)/2&3/[email protected]𝟐−𝟐(𝟏)&2−2((−3)/2)&3−2(3/2)@3&−2&2)] = [■8(1/2&0&[email protected]−2(1/2)&1−2(0)&0−2(0)@0&0&1)] A [■8(1&(−3)/2&3/[email protected]𝟎&5&[email protected]&−2&2)] = [■8(1/2&0&[email protected]−1&1&[email protected]&0&1)] A R3 → R3 – 3R1 [■8(1&(−3)/2&3/[email protected]&5&[email protected]𝟑−𝟑(𝟏)&−2−3((−3)/2)&2−3(3/2) )]= [■8(1/2&0&[email protected]−1&1&[email protected]−3(1/2)&0−3(0)&1−3(0))] A [■8(1&(−3)/2&3/[email protected]&5&[email protected]𝟎&5/2&(−5)/2)]= [■8(1/2&0&[email protected]−1&1&[email protected](−3)/2&0&1)] A R2 →R_2/5 , [■8(1&(−3)/2&3/[email protected]&𝟓/𝟓&[email protected]&5/2&(−5)/2)] = [■8(1/2&0&[email protected](−1)/5&1/5&[email protected](−3)/2&0&1)] A [■8(1&(−3)/2&3/[email protected]&𝟏&[email protected]&5/2&(−5)/2)] = [■8(1/2&0&[email protected](−1)/5&1/5&[email protected](−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&[email protected]&5/2&(−5)/2)]= [■8(1/2+3/2 ((−1)/5)&0+3/2 (1/5)&0+3/2(0)@(−1)/5&1/5&[email protected](−3)/2&0&1)] A [■8(1&𝟎&3/[email protected]&1&[email protected]&5/2&(−5)/2)]= [■8(2/10&3/10&[email protected](−1)/5&1/5&[email protected](−3)/2&0&1)] A R3 → R3 – 5/2 R2 [■8(1&0&3/[email protected]&1&[email protected]−5/2(0)&𝟓/𝟐−𝟓/𝟐(𝟏)&(−5)/2−(−5)/2(0))]= [■8(1/5&3/10&[email protected](−1)/5&1/5&[email protected](−3)/2−5/2 ((−1)/5)&0−5/2 (1/5)&1−5/2(0))] A [■8(1&0&3/[email protected]&1&[email protected]&𝟎&(−5)/2)] = [■8(1/5&3/10&[email protected](−1)/5&1/5&[email protected]−1&(−1)/2&1)] A R3 → (−2)/5 R3 [■8(1&0&3/[email protected]&1&[email protected]((−2)/5)&0(−2/5)&(−𝟓)/𝟐 ((−𝟐)/𝟓) )]= [■8(1/5&3/10&[email protected](−1)/5&1/5&[email protected]−1((−2)/5)&(−1)/2 ((−2)/5)&1((−2)/5) )] A [■8(1&0&3/[email protected]&1&[email protected]&0&𝟏)] = [■8(1/5&3/10&[email protected](−1)/5&1/5&[email protected]/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&[email protected]&0&1)]= [■8(1/5−3/2 (2/5)&3/10−3/2 (1/5)&0−3/2 ((−2)/5)@(−1)/5&1/5&[email protected]/5&1/5&(−2)/5)] A [■8(1&0&𝟎@0&1&[email protected]&0&1)] = [■8((−2)/5&0&3/[email protected](−1)/5&1/5&[email protected]/5&1/5&(−2)/5)] A I = [■8((−2)/5&0&3/[email protected](−1)/5&1/5&[email protected]/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)]