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Ex 3.4, 17 - Find inverse [2 0 -1 5 1 0 0 1 3] - NCERT - Ex 3.4

  1. Class 12
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Ex3.4, 17 Find the inverse of each of the matrices, if it exists. [■8(2&0&−1@5&1&0@0&1&3)] Let A =[■8(2&0&−1@5&1&0@0&1&3)] We know that A = IA [■8(2&0&−1@5&1&0@0&1&3)]= [■8(1&0&0@0&1&0@0&0&1)] A R1 →1/2 R1 , [■8(𝟐/𝟐&0/2&(−1)/2@5&1&0@0&1&3)]= [■8(1/2&0/2&0/2@0&1&0@0&0&1)] A [■8(𝟏&0&−1/2@5&1&0@0&1&3)] = [■8(1/2&0&0@0&1&0@0&0&1)] A R2 → R2 – 5R1 [■8(1&0&−1/2@𝟓−𝟓(𝟏)&1−5(0)&0−5((−1)/2)@0&1&3)] = [■8(1/2&0&0@0−5(1/2)&1−5(0)&0−5(0)@0&0&1)] A [■8(1&0&−1/2@𝟎&1&5/2@0&0&3)]= [■8(1/2&0&0@−5/2&1&0@0&0&1)] A R3 → R3 – R2 [■8(1&0&−1/2@0&1&5/2@𝟎&0&3−5/2)]= [■8(1/2&0&0@−5/2&1&0@0−(−5/2)&0−1&1−0)] A [■8(1&0&−1/2@0&1&5/2@𝟎&0&1/2)]= [■8(1/2&0&0@−5/2&1&0@5/2&−1&1)] A R3 → 2R3 [■8(1&0&−1/2@0&1&5/2@2×0&2×0&𝟐×𝟏/𝟐)]= [■8(1/2&0&0@−5/2&1&0@2×5/2&2×(−1)&2(1))] A [■8(1&0&−1/2@0&1&5/2@0&0&𝟏)]= [■8(1/2&0&0@−5/2&1&0@5&−2&2)] A R1 → R1 + 1/2 R3 [■8(1+1/2(0)&0+1/2(0)&(−𝟏)/𝟐+𝟏/𝟐(𝟏)@0&1&5/2@0&0&1)]= [■8(1/2+1/2(5)&0+1/2(−2)&0+1/2(2)@(−5)/2&1&0@5&−2&2)] A [■8(1&0&𝟎@0&1&5/2@0&0&1)]= [■8(3&−1&1@(−5)/2&1&0@5&−2&2)] A R2 → R2 − 5/2 R3 [■8(1&0&0@0−5/2(0)&1−5/2(0)&𝟓/𝟐−𝟓/𝟐(𝟏)@0&0&1)]= [■8(3&−1&1@(−5)/2−5/2 (5)&1−5/2(−2)&0−5/2@5&−2&2)] A [■8(1&0&0@0&1&𝟎@0&0&1)]= [■8(−3&−1&1@(−30)/2&6&−5@5&−2&2)] A "I"= [■8(−3&−1&1@−15&6&−5@5&−2&2)] A This is similar to I = A-1A Thus, A-1 = [■8(3&−1&1@−15&6&−5@5&−2&2)]

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