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Example 24 - Obtain inverse using elementary operations - Examples

Example 24 - Chapter 3 Class 12 Matrices - Part 2
Example 24 - Chapter 3 Class 12 Matrices - Part 3 Example 24 - Chapter 3 Class 12 Matrices - Part 4 Example 24 - Chapter 3 Class 12 Matrices - Part 5

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Example 24 Obtain the inverse of the following matrix using elementary operations A = [■8(0&1&[email protected]&2&[email protected]&1&1)] Given A = [■8(0&1&[email protected]&2&[email protected]&1&1)] We know that A = IA [■8(0&1&[email protected]&2&[email protected]&1&1)] = [■8(1&0&[email protected]&1&[email protected]&0&1)] A R1↔R2 [■8(𝟏&2&[email protected]&1&[email protected]&1&1)] = [■8(0&1&[email protected]&0&[email protected]&0&1)] A R3 → R3 – 3R1 [■8(1&2&[email protected]&1&[email protected]𝟑−𝟑(𝟏)&1−3(2)&1−3(3))] = [■8(0&1&[email protected]&0&[email protected]−3(0)&0−3(1)&1−3(0))]A [■8(1&2&[email protected]&1&[email protected]𝟎&−5&−8)] = [■8(0&1&[email protected]&0&[email protected]&−3&1)] R1 → R1 – 2R2 [■8(1−2(0)&𝟐−𝟐(𝟏)&3−2(2)@0&1&[email protected]&−5&−8)] = [■8(0−2(1)&1−2(0)&0−2(0)@1&0&[email protected]&−3&1)]A [■8(1&𝟎&−[email protected]&1&[email protected]&−5&−8)] = [■8(−2&1&[email protected]&0&[email protected]&−3&1)] A R3 → R3 + 5R2 [■8(1&0&−[email protected]&1&[email protected]+5(0)&−𝟓+𝟓(𝟏)&−8+5(2))] = [■8(−2&1&[email protected]&0&[email protected]+5(1)&−3+5(0)&1+5(0))] A [■8(1&0&−[email protected]&1&[email protected]&𝟎&2)] = [■8(−2&1&[email protected]&0&[email protected]&−3&1)] A R3 → 1/2 R3 [■8(1&0&−[email protected]&1&[email protected]/2&0/2&𝟐/𝟐)] = [■8(−2&1&[email protected]&0&[email protected]/2&(−3)/2&1/2)] A R1 → R1 + R3 [■8(1+0&0+0&−𝟏+𝟏@0&1&[email protected]&0&1)]=[■8(−2+5/2&1+((−3)/2)&0+1/[email protected]&0&[email protected]/2&(−3)/2&1/2)] A [■8(1&0&𝟎@0&1&[email protected]&0&1)] = [■8(1/2&(−1)/2&1/[email protected]&0&[email protected]/2&(−3)/2&1/2)] A R2 → R2 – 2R3 [■8(1&0&[email protected]−2(0)&1−2(0)&𝟐−𝟐(𝟏)@0&0&1)] = [■8(1/2&(−1)/2&1/[email protected]−2(5/2)&0−2((−3)/2)&0−2(1/2)@5/2&(−3)/2&1/2)]A [■8(1&0&[email protected]&1&𝟎@0&0&1)] = [■8(1/2&(−1)/2&1/[email protected]−4&3&−[email protected]/2&(−3)/2&1/2)] A I= [■8(1/2&(−1)/2&1/[email protected]−4&3&−[email protected]/2&(−3)/2&1/2)] A This is similar to I = A-1 A Hence, A-1 = [■8(𝟏/𝟐&(−𝟏)/𝟐&𝟏/𝟐@−𝟒&𝟑&−𝟏@𝟓/𝟐&(−𝟑)/𝟐&𝟏/𝟐)]

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.