Ex 4.5, 15 - Chapter 4 Class 12 Determinants

Transcript

Ex 4.5, 15 For the matrix A = 1 1 1 1 2 3 2 1 3 show that A3 6A2 + 5A + 11I = O. Hence, find A 1. Calculating A2 A2 = A.A = 1 1 1 1 2 3 2 1 3 1 1 1 1 2 3 2 1 3 = 1 1 +1 1 +1(2) 1 1 +1 2 +1( 1) 1 1 +1 3 +1(3) 1 1 +2 1 +( 3)(2) 1 1 +2 2 +( 3)( 1) 1 1 +2 3 +( 3)(3) 2 1 +( 1) 1 +3(2) 2 1 +( 1) 2 +3( 1) 2 1 + 1 3 +3(3) = 1+1+2 1+2 1 1 3+3 1+2 6 1+4+3 1 6 9 2 1+6 2 2 3 2+3+9 = 4 2 1 3 8 14 7 3 14 Now finding A3 A3 = A2 A A3 = 4 2 1 3 8 14 7 3 14 1 1 1 1 2 3 2 1 3 = 4 1 +2 1 +1(2) 4 1 +2 2 +1( 1) 4 1 +2 3 +1(3) 3 1 +8 1 +( 4)(2) 3 1 +8 2 +( 4)( 1) 3 1 +8 3 +( 4)(3) 7 1 +( 3) 1 +14(2) 7 1 +( 3) 2 +14( 1) 7 1 + 3 3 +14(3) = 4+2+2 4+4 1 4 6+3 3+8 8 3+16 4 3 24 42 7 3+28 7 6 14 7+9+42 = 8 7 1 23 27 69 32 13 58 Now Putting value of A3 , A2 in A3 6A2 + 5A + 11I = 8 7 1 23 27 69 32 13 58 6 4 2 1 3 8 14 7 3 14 + 5 1 1 1 1 2 3 2 1 3 + 11 1 0 0 0 1 0 0 0 1 = 8 7 1 23 27 69 32 13 58 6(4) 6(2) 6(1) 6( 3) 6(8) 6( 14) 6(7) 6( 3) 6(14) + 5(1) 5(1) 5(1) 5 1 5(2) 5( 3) 5(2) 5( 1) 5(3) + 11(1) 0 0 0 11(1) 0 0 0 11(1) = 8 7 1 23 27 69 32 13 58 24 12 6 18 48 84 42 18 84 + 5 5 5 5 10 15 10 5 15 + 11 0 0 0 11 0 0 0 11 = 8 24+5+11 7 12+5+0 1 6+5+0 23+18+5+0 27 48+10+11 69+84 15+0 32 42+10+0 13+18 5+0 58+84+15+11 = 24 24 12 12 6 6 23+23 48+48 84 84 42+42 18 18 84 84 = 0 0 0 0 0 0 0 0 0 = O Hence proved Finding A-1 A3 6A2 + 5A + 11I = O Post multiplying A-1 both sides (A3 6A2 + 5A +11I)A-1 = O A-1 A3 .A-1 6 A2. A-1 + 5AA-1 + 11IA-1 = O A2. A.A-1 6A.AA-1 + 5AA-1 + 11A-1 = O A2 (AA-1) 6A(AA-1) + 5(AA-1)11A-1 = O A2 I 6AI + 5I + 11A-1 = 0 A2 6A + 5I + 11A-1 = 0 11A-1 = A2 + 6A 5I A-1 = 1 11 ( A2 + 6A 5I) Putting values A-1 = 1 11 4 2 1 3 8 14 7 3 14 + 6 1 1 1 1 2 3 2 1 3 5 1 0 0 0 1 0 0 0 1 = 1 11 4 2 1 3 8 14 7 3 14 + 6(1) 6(1) 6(1) 6(1) 6(2) 6( 3) (2) 6( 1) 6(3) + 5 5(1) 0 0 0 5(1) 0 0 0 5(1) = 1 11 4 2 1 3 8 14 7 3 14 + 6 6 6 6 12 18 12 6 18 5 0 0 0 5 0 0 0 5 = 1 11 4+6 5 2+6+0 1+6+0 3+6+0 8+12 5 14 18+0 7+12+0 3 6+0 14+18 5 = 1 11 3 4 5 9 1 4 5 3 1 Thus, A-1 = 1 11 3 4 5 9 1 4 5 3 1