Examples

Chapter 4 Class 12 Determinants
Serial order wise

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Example 26 Show that the matrix A = [■8(2&[email protected]&2)] satisfies the equation A2 – 4A + I = O, where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A–1. First calculating A2 A2 = A. A = [■8(2&[email protected]&2)] [■8(2&[email protected]&2)] = [■8(2(2)+3(1)&2(3)+3(2)@1(2)+2(1)&1(3)+2(2))] = [■8(4+3&[email protected]+2&3+4)] = [■8(7&[email protected]&7)] Now, solving A2 – 4A + I Putting values = [■8(7&[email protected]&7)] – 4 [■8(2&[email protected]&2)] + [■8(1&[email protected]&1)] = [■8(7&12[email protected]&7)] – [■8(4(2)&4(3)@4(1)&4(2))] + [■8(1&[email protected]&1)] = [■8(7&[email protected]&7)] – [■8(8&[email protected]&8)] + [■8(1&[email protected]&1)] = [■8(7−8+1&12−[email protected]−4+0&7−8+1)] = [■8(8−8&12−[email protected]−4&8−8)] = [■8(0&[email protected]&0)] = O Thus, A2 – 4A + I = O Hence proved Now, Finding A-1 using equation A2 – 4A + I = O Post multiplying by A-1 both sides (A2 – 4A +I) A-1 = OA-1 A2 . A-1 – AA-1 + I. A-1 = O A . (AA-1) – 4AA-1 + IA-1 = O A. I – 4I + IA-1 = O A – 4I + A-1 = O A-1 = O − A + 4I A-1 = −A + 4I (OA–1 = O) (AA-1 = I) (IA-1 = A-1) Putting values A-1 = –[■8(2&[email protected]&2)] + 4 [■8(1&[email protected]&1)] = [■8(−2&−[email protected]−1&−2)] + [■8(4&[email protected]&4)] = [■8(−2+4&−[email protected]−1+0&−2+4)] = [■8(2&−[email protected]−1&2)] Thus, A-1 = [■8(𝟐&−𝟑@−𝟏&𝟐)]

#### 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.