Check sibling questions

Example 26 - Show that matrix A = [2 3 1 2] satisfies equation A2 - 4A

Example 26 - Chapter 4 Class 12 Determinants - Part 2
Example 26 - Chapter 4 Class 12 Determinants - Part 3
Example 26 - Chapter 4 Class 12 Determinants - Part 4

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Transcript

Example 26 Show that the matrix A = [■8(2&3@1&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&3@1&2)] [■8(2&3@1&2)] = [■8(2(2)+3(1)&2(3)+3(2)@1(2)+2(1)&1(3)+2(2))] = [■8(4+3&6+6@2+2&3+4)] = [■8(7&12@4&7)] Now, solving A2 – 4A + I Putting values = [■8(7&12@4&7)] – 4 [■8(2&3@1&2)] + [■8(1&0@0&1)] = [■8(7&12@4&7)] – [■8(4(2)&4(3)@4(1)&4(2))] + [■8(1&0@0&1)] = [■8(7&12@4&7)] – [■8(8&12@4&8)] + [■8(1&0@0&1)] = [■8(7−8+1&12−12+0@4−4+0&7−8+1)] = [■8(8−8&12−12@4−4&8−8)] = [■8(0&0@0&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&3@1&2)] + 4 [■8(1&0@0&1)] = [■8(−2&−3@−1&−2)] + [■8(4&0@0&4)] = [■8(−2+4&−3+0@−1+0&−2+4)] = [■8(2&−3@−1&2)] Thus, A-1 = [■8(𝟐&−𝟑@−𝟏&𝟐)]

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.