Example 26 - Chapter 4 Class 12 Determinants (Term 1)

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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)]