Example 14 - 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 14 If A = [■8(2&3@1&−4)] and B = [■8(1&−2@−1&3)] , then verify that (AB)-1 = B-1 A-1 Solving L.H.S (AB) –1 First calculating AB AB = [■8(2&3@1&−4)] [■8(1&−2@−1&3)] = [■8(2(1)+3(−1)&2(−2)+3(3)@1(1)+( −4)(−1)&1(−2)+(⤶7−4)3)] = [■8(2−3&−4+9@1+4&−2−12)] = [■8(−𝟏&𝟓@𝟓&−𝟏𝟒)] Now, (AB)-1 = 1/(|AB|) adj (AB) exists if |AB| ≠ 0 |AB| = |■8(−1&5@5&−14)| = (-1)(-14) – 5(5) = 14 – 25 = –11 Since |AB| ≠ 0, (AB)-1 exists AB = [■8(−1&5@5&−14)] adj (AB) = [■8(−1&5@5&−14)] = [■8(−𝟏𝟒&−𝟓@−𝟓&−𝟏)] Now, (AB)–1 = 1/(|AB|) adj (AB) Putting values = 1/(−11) [■8(−14&−5@−5&−1)] = 𝟏/𝟏𝟏 [■8(𝟏𝟒&𝟓@𝟓&𝟏)] Solving R.H.S B-1A-1 First Calculating B–1 B = [■8(1&−2@−1&3)] B = 1/(|B|) adj (B) exists if |B| ≠ 0 |B| = |■8(1&−2@−1&3)| = 3 – 2 = 1 Since |B| ≠ 0 , B-1 exist B = [■8(1&−2@−1&3)] adj (B) =[■8(1&−2@−1&3)] = [■8(3&2@1&1)] Now, (B)–1 = 1/(|B|) adj (B) Putting values = 1/1 [■8(3&2@1&1)] = [■8(𝟑&𝟐@𝟏&𝟏)] Finding A-1 A-1 = 1/(|A|) adj (A) exists if |A| ≠ 0 |A| = |■8(2&3@1&−4)| = 2 ( – 4) – 1( 3) = – 8 – 3 = – 11 Since |A| ≠ 0 , A–1 exists A = [■8(2&3@1&−4)] adj (A) = [■8(2&3@1&−4)] = [■8(−4&−3@−1&2)] Now, A-1 = 1/(|A|) adj (A) = 1/(−11) [■8(−4&−3@−1&2)] = 𝟏/𝟏𝟏 [■8(𝟒&𝟑@𝟏&−𝟐)] Thus, B-1A-1 = [■8(3&2@1&1)] × 1/11 [■8(4&3@1&−2)] = 𝟏/𝟏𝟏 [■8(𝟑&𝟐@𝟏&𝟏)] [■8(𝟒&𝟑@𝟏&−𝟐)] = 1/11 [■8(3(4)+2(1)&3(3)+2(−2)@ 1(4)+1(1)&1(3)+1(−2))] = 1/11 [■8(12+2&9−4@4+1&3−2)] = 𝟏/𝟏𝟏 [■8(𝟏𝟒&𝟓@𝟓&𝟏)] = L.H.S ∴ L.H.S = R.H.S Hence proved