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Question 24 (OR 2
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Find the inverse of the following matrix using elementary transformations

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

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important

Question 13 (Or 2nd)

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important You are here

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd)

Question 27 (Or 1st) Important

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 24, 2021 by Teachoo

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Question 24 (OR 2
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nd
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question)
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Find the inverse of the following matrix using elementary transformations

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

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)]