Examples

Chapter 3 Class 12 Matrices
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

Transcript

Example 18 If A = [■8(1&2&[email protected]&−2&[email protected]&2&1)] then show that A3 – 23A – 40I = O Finding A2 A2 = AA = [■8(1&2&[email protected]&−2&[email protected]&2&1)] [■8(1&2&[email protected]&−2&[email protected]&2&1)] = [■8(1(1)+2(3)+3(4)&1(2)+2(−2)+3(2)&1(3)+2(1)+3(1)@3(1)+(−2)(3)+1(4)&3(2)+(−2)(−2)+1(2)&3(3)+(−2)(1)+1(1)@4(1)+2(3)+1(4)&4(2)+2(−2)+1(2)&4(3)+(2)(1)+1(1))] = [■8(1+6+12&2−4+6&[email protected]−6+4&6+4+2&9−[email protected]+6+4&8−4+2&12+2+1)] = [■8(19&4&[email protected]&12&[email protected]&6&15)] Finding A3 A3 = A2 A = [■8(19&4&[email protected]&12&[email protected]&6&15)] [■8(1&2&[email protected]&−2&[email protected]&2&1)] = [■8(19(1)+4(3)+8(4)&19(2)+4(−2)+8(2)&19(3)+4(1)+8(1)@1(1)+12(3)+8(4)&1(2)+12(−2)+8(2)&1(3)+12(1)+8(1)@14(1)+6(3)+15(4)&14(2)+6(−2)+15(2)&14(3)+6(1)+15 (1))] = [■8(19+12+32&38−8+16&[email protected]+36+32&2−24+16&[email protected]+18+60&28−12+30&42+6+15)] = [■8(63&46&[email protected]&−6&[email protected]&46&63)] Calculating A3 – 23A – 40I = [■8(63&46&[email protected]&−6&[email protected]&46&63)] −23 [■8(1&2&[email protected]&−2&[email protected]&2&1)] −40 [■8(1&0&[email protected]&1&[email protected]&0&1)] = [■8(63&46&[email protected]&−6&[email protected]&46&63)] −[■8(23×1&23×2&23×[email protected]×3&23×(−2)&23×[email protected]×4&23×(2)&23×1)] − [■8(1×40&0×40&0×[email protected]×40&1×40&0×[email protected]×40&0×40&1×40)] = [■8(63&46&[email protected]&−6&[email protected]&46&63)] ⤶7− [■8(23&46&[email protected]&−46&[email protected]&46&23)] ⤶7− [■8(40&0&[email protected]&40&[email protected]&0&40)] = [■8(63−23−40&46−46+0&69−[email protected]−69+0&−6+46−40&23−[email protected]−92+0&46−46+0&63−23−40)] = [■8(0&0&[email protected]&0&[email protected]&0&0)] = O Hence proved.