Example 15 - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Finding inverse when Equation of matrice given
Finding inverse when Equation of matrice given
Last updated at April 16, 2024 by Teachoo
Example 15 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 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)]