Ex 3.4

Chapter 3 Class 12 Matrices
Serial order wise

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### Transcript

Ex3.4, 17 Find the inverse of each of the matrices, if it exists. [■8(2&0&−[email protected]&1&[email protected]&1&3)] Let A =[■8(2&0&−[email protected]&1&[email protected]&1&3)] We know that A = IA [■8(2&0&−[email protected]&1&[email protected]&1&3)]= [■8(1&0&[email protected]&1&[email protected]&0&1)] A R1 →1/2 R1 , [■8(𝟐/𝟐&0/2&(−1)/[email protected]&1&[email protected]&1&3)]= [■8(1/2&0/2&0/[email protected]&1&[email protected]&0&1)] A [■8(𝟏&0&−1/[email protected]&1&[email protected]&1&3)] = [■8(1/2&0&[email protected]&1&[email protected]&0&1)] A R2 → R2 – 5R1 [■8(1&0&−1/[email protected]𝟓−𝟓(𝟏)&1−5(0)&0−5((−1)/2)@0&1&3)] = [■8(1/2&0&[email protected]−5(1/2)&1−5(0)&0−5(0)@0&0&1)] A [■8(1&0&−1/[email protected]𝟎&1&5/[email protected]&0&3)]= [■8(1/2&0&[email protected]−5/2&1&[email protected]&0&1)] A R3 → R3 – R2 [■8(1&0&−1/[email protected]&1&5/[email protected]𝟎&0&3−5/2)]= [■8(1/2&0&[email protected]−5/2&1&[email protected]−(−5/2)&0−1&1−0)] A [■8(1&0&−1/[email protected]&1&5/[email protected]𝟎&0&1/2)]= [■8(1/2&0&[email protected]−5/2&1&[email protected]/2&−1&1)] A R3 → 2R3 [■8(1&0&−1/[email protected]&1&5/[email protected] × 0&2 × 0&𝟐 × 𝟏/𝟐)] = [■8(1/2&0&[email protected]−5/2&1&[email protected] × 5/2&2 × (−1)&2 × 1)] A [■8(1&0&−1/[email protected]&1&5/[email protected]&0&𝟏)]= [■8(1/2&0&[email protected]−5/2&1&[email protected]&−2&2)] A R1 → R1 + 1/2 R3 [■8(1+1/2(0)&0+1/2(0)&(−𝟏)/𝟐+𝟏/𝟐(𝟏)@0&1&5/[email protected]&0&1)]= [■8(1/2+1/2(5)&0+1/2(−2)&0+1/2(2)@(−5)/2&1&[email protected]&−2&2)] A [■8(1&0&𝟎@0&1&5/[email protected]&0&1)]= [■8(3&−1&[email protected](−5)/2&1&[email protected]&−2&2)] A R2 → R2 − 5/2 R3 [■8(1&0&[email protected]−5/2(0)&1−5/2(0)&𝟓/𝟐−𝟓/𝟐(𝟏)@0&0&1)] = [■8(3&−1&[email protected](−5)/2−5/2 (5)&1−5/2 (−2)&0−5/[email protected]&−2&2) (2)] A [■8(1&0&[email protected]&1&𝟎@0&0&1)]= [■8(3&−1&[email protected](−30)/2&6&−[email protected]&−2&2)] A "I" = [■8(3&−1&[email protected]−15&6&−[email protected]&−2&2)] A This is similar to I = A-1A Thus, A-1 = [■8(𝟑&−𝟏&𝟏@−𝟏𝟓&𝟔&−𝟓@𝟓&−𝟐&𝟐)]