Examples

Chapter 4 Class 12 Determinants
Serial order wise

### Transcript

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

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.