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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.