Question 24 (OR 2 nd question)

Find the inverse of the following matrix using elementary transformations

[2 -1 3 -5 3 1 -3 2 3]

Find inverse of matrix using elementary transformation - Teachoo

Question 24 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
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Question 24 (OR 2nd question) Find the inverse of the following matrix using elementary transformations [■8(2&−1&3@−5&3&1@−3&2&3)] Let A = [■8(2&−1&3@−5&3&1@−3&2&3)] We know that A = IA [■8(2&−1&3@−5&3&1@−3&2&3)]= [■8(1&0&0@0&1&0@0&0&1)] A R1 → R1 + 𝑹_𝟐/𝟓 [■8(1&(−2)/5&16/5@−5&3&1@−3&2&3)]= [■8(1&1/5&0@0&1&0@0&0&1)] A R2 → R2 + 5R1 [■8(1&(−2)/5&16/5@−5+5(1)&3+5((−2)/5)&1+5(16/5)@−3&2&3)]= [■8(1&1/5&0@0+5(1)&1+5(1/5)&0+5(0)@0&0&1)] A [■8(1&(−2)/5&16/5@0&1&17@−3&2&3)]= [■8(1&1/5&0@5&2&0@0&0&1)] A R3 → R3 + 3R1 [■8(1&(−2)/5&16/5@0&1&17@−3+3(1)&2+3(−2/5)&3+3(16/5) )]= [■8(1&1/5&0@5&2&0@0+3(1)&0+3(1/5)&1+3(0))] A [■8(1&(−2)/5&16/5@0&1&17@0&4/5&63/5)]= [■8(1&1/5&0@5&2&0@3&3/5&1)] A R1 → R1 + 𝟐/𝟓R2 [■8(1+2/5(0)&(−2)/5+2/5(1)&16/5+2/5(17)@0&1&17@0&4/5&63/5)]= [■8(1+2/5(5)&1/5+2/5(2)&0+2/5(0)@5&2&0@3&3/5&1)] A [■8(1&0&10@0&1&17@0&4/5&63/5)]= [■8(3&1&0@5&2&0@3&3/5&1)] A R3 → R3 – 𝟒/𝟓R2 [■8(1&0&10@0&1&17@0−4/5(0)&4/5−4/5(1)&63/5−4/5(17))]= [■8(3&1&0@5&2&0@3−4/5(5)&3/5−4/5(2)&1−4/5(0))] A [■8(1&0&10@0&1&17@0&0&−1)]= [■8(3&1&0@5&2&0@−1&−1&1)] A R3 → R3 × –1 [■8(1&0&10@0&1&17@0&0&1)]= [■8(3&1&0@5&2&0@1&1&−1)] A R1 → R1 – 10R3 [■8(1−10(0)&0−10(0)&10−10(1)@0&1&17@0&0&1)]= [■8(3−10(1)&1−10(1)&0−10(−1)@5&2&0@1&1&−1)] A [■8(1&0&0@0&1&17@0&0&1)]= [■8(−7&−9&10@5&2&0@1&1&−1)] A R2 → R2 – 17R3 [■8(1&0&0@0−17(0)&1−17(0)&17−17(1)@0&0&1)]= [■8(−7&−9&10@5−17(1)&2−17(1)&0−17(−1)@1&1&−1)] A [■8(1&0&0@0&1&0@0&0&1)] = [■8(−7&−9&10@−12&−15&17@1&1&−1)] A "I"= [■8(−7&−9&10@−12&−15&17@1&1&−1)] A This is similar to I = A-1A Thus, A-1 = [■8(−7&−9&10@−12&−15&17@1&1&−1)]

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.