Example 13 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Inverse of two matrices and verifying properties
Inverse of two matrices and verifying properties
Last updated at April 16, 2024 by Teachoo
Example 13 If A = [■8(1&3&3@1&4&3@1&3&4)], then verify that A adj A = |A| I. Also find A–1. Solving L.H.S A (adj A) First Calculating adj A adj A = [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] Now, A = [■8(1&3&3@1&4&3@1&3&4)] M11 = |■8(4&3@3&4)| = 4(4) – 3(3) = 7 M12 = |■8(1&3@1&4)| = 1(4) – 1(3) = 1 M13 = |■8(1&4@1&3)| = 1(3) – 1(4) = –1 M21 = |■8(3&3@3&4)| = 3(4) – 3(3) = 3 M22 = |■8(1&3@1&4)| = 1(4) – 1(3) = 1 M23 = |■8(1&3@1&3)| = 1(3) – 1(3) = 0 M31 = |■8(3&3@4&3)| = 3(3) – 4(3) = – 3 M32 = |■8(1&3@1&3)| = 1(3) – 1(3) = 0 M33 = |■8(1&3@1&4)| = 1(4) – 1(3) = 1 Now, A11 = (–1)1 + 1 M11 = (–1)2 7 = 7 A12 = (–1)1+2 M12 = (–1)3 (1) = –1 A13 = (–1)1+3 M13 = (–1)4 (–1) = –1 A21 = (–1)2+1 M21 = (–1)3 (3) = –3 A22 = (–1)2+2 M22 = (–1)4 (1) = 1 A23 = (–1)2+3 M23 = (–1)5 0 = 0 A31 = (–1)3+1 M31 = (–1)4 (– 3) = –3 A32 = (–1)3+2 M32 = (–1)5 0 = 0 A33 = (–1)3+3 M33 = (–1)6 (1) = 1 Thus, adj (A) = [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] = [■8(7&−3&−3@−1&1&0@−1&0&1)] Finding A (adj A) A adj (A) = [■8(1&3&3@1&4&3@1&3&4)] [■8(7&3&−3@−1&1&0@−1&0&1)] = [■8(1(7)+3(⤶7−1)+3(−1)&1(−3)+3(1)+3(0)&1(−3)+3(0)+3(1)@1(7)+4(⤶7−1)+3(−1)&1(−3)+4(1)+3(0)&1(−3)+4(0)+3(1)@1(7)+3(⤶7−1)+4(−1)&1(−3)+3(1)+4(0)&1(−3)+3(0)+4(1))] = [■8(7−3−3&−3+3+0&−3+0+3@7−4−3&−3+4+0&−3+0+3@7−3−4&−3+3+0&−3+0+4)] = [■8(𝟏&𝟎&𝟎@𝟎&𝟏&𝟎@𝟎&𝟎&𝟏)] Solving R.H.S |A| I Calculating |A| |A| = |■8(1&3&3@1&4&3@1&3&4)| = 1 (4(4) – 3(3)) – 3(1(4) – 1(3)) + 3(1(4) – 1(3)) = 1(7) – 3(1) +3( – 1) = 7 – 3 – 3 = 1 Now, |A| I = 1 [■8(1&0&0@0&1&0@0&0&1)] = [■8(1&0&0@0&1&0@0&0&1)] = L.H.S Thus, A(adj A) = |A| I Hence proved Finding A-1 We know that A-1 = 1/(|A|) (adj A) exists if |A| ≠ 0 Here, |A| = 1 ≠ 0 Thus A-1 exists So, A-1 = 1/(|A|) (adj A) = 1/1 [■8(7&−3&−3@−1&1&0@−1&0&1)] = [■8(𝟕&−𝟑&−𝟑@−𝟏&𝟏&𝟎@−𝟏&𝟎&𝟏)]