# Ex 4.5, 9 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

Last updated at Jan. 23, 2020 by Teachoo

Transcript

Ex 4.5, 9 Find the inverse of each of the matrices (if it exists). [■8(2&1&3@4&−1&0@−7&2&1)] Let A = [■8(2&1&3@4&−1&0@−7&2&1)] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Calculating |A| |A| = |■8(2&1&3@4&−1&0@−7&2&1)| = 2 |■8(–1&0@2&1)| – 1 |■8(4&0@−7&1)| + 3 |■8(4&−1@−7&2)| = 2(–1 – 0) – 1 (4 – 0) + 3 (8 – 7) = 2 (–1) – 1 (4) + 3 (1) = –3 Since |𝐴| ≠ 0 , A–1 exists Calculating adj (A) adj (A) = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(2&1&3@4&−1&0@−7&2&1)] M11 = |■8(−1&0@2&1)| = –1(1) – 0(2) = -1 M12 = |■8(4&0@−7&1)| = 4(1) – 0(–7) = 4 M13 = |■8(4&−1@−7&2)| = 4(2) – (-7)(-1) = 1 M21 = |■8(1&3@2&1)| = 1(1) – 2(3) = – 5 M22 = |■8(2&3@−7&1)| = 2(1) – (−7)(3)= 23 M23 = |■8(2&1@−7&2)| = 2(2) – (−7)(1) = 11 M31 = |■8(1&3@−1&0)| = 1(0) – (−1)(3) = 3 M32 = |■8(2&3@4&0)| = 2(0) – 4(3) = –12 M33 = |■8(2&1@4&−1)| = 2(-1) – 4(1) = – 6 Now, A11 = (–1)1 + 1 M11 = (–1)2 (–1) = 1 (–1) = – 1 A12 = (–1)1+2 M12 = (–1)3 ( 4) = ( –1) (4) = –4 A13 = (–1)1+3 M13 = (–1)4 (1) = 1 A21 = (–1)2+1 M21 = (–1)3 ( –5) = (−1)(−5) = 5 A22 = (–1)2+2 M22 = (–1)4 ( –23) = 23 A23 = (–1)2+3 M23 = (–1)5 11 = −1 (11) = – 11 A31 = (–1)3+1 M31 = (–1)4 (4) = 1 (3) = 3 A32 = (–1)3+2 M32 = (–1)5 (–12) = (–1) (–12) = 12 A33 = (–1)3+3 M33 = (–1)6 (–6) = 1 . (–6) = – 6 Thus, adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [■8(−1&5&3@−4&23&12@1&−11&−6)] Calculating inverse A– 1 = 1/(|A|) ( adj (A)) = 1/(−3) [■8(−1&5&3@−4&23&12@1&−11&−6)] = (−𝟏)/𝟑 [■8(−𝟏&𝟓&𝟑@−𝟒&𝟐𝟑&𝟏𝟐@𝟏&−𝟏𝟏&−𝟔)]

Ex 4.5

Ex 4.5, 1

Ex 4.5, 2

Ex 4.5, 3 Important

Ex 4.5, 4 Important

Ex 4.5, 5

Ex 4.5, 6 Important

Ex 4.5, 7

Ex 4.5, 8

Ex 4.5, 9 You are here

Ex 4.5, 10 Important

Ex 4.5, 11 Important

Ex 4.5, 12

Ex 4.5, 13

Ex 4.5, 14 Important

Ex 4.5, 15 Important

Ex 4.5, 16

Ex 4.5, 17 (MCQ) Important

Ex 4.5, 18 (MCQ) Important

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.