Solving Equation

Chapter 3 Class 12 Matrices
Concept wise

### Transcript

Example 18 If A = [■8(1&2&3@3&−2&1@4&2&1)] then show that A3 – 23A – 40I = O Finding A2 A2 = AA = [■8(1&2&3@3&−2&1@4&2&1)] [■8(1&2&3@3&−2&1@4&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&3+2+3@3−6+4&6+4+2&9−2+1@4+6+4&8−4+2&12+2+1)] = [■8(𝟏𝟗&𝟒&𝟖@𝟏&𝟏𝟐&𝟖@𝟏𝟒&𝟔&𝟏𝟓)] Finding A3 A3 = A2 A = [■8(19&4&8@1&12&8@14&6&15)] [■8(1&2&3@3&−2&1@4&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&57+4+8@1+36+32&2−24+16&3+12+8@14+18+60&28−12+30&42+6+15)] = [■8(𝟔𝟑&𝟒𝟔&𝟔𝟗@𝟔𝟗&−𝟔&𝟐𝟑@𝟗𝟐&𝟒𝟔&𝟔𝟑)] Calculating A3 – 23A – 40I = [■8(63&46&69@69&−6&23@92&46&63)] −23 [■8(1&2&3@3&−2&1@4&2&1)] −40 [■8(1&0&0@0&1&0@0&0&1)] = [■8(63&46&69@69&−6&23@92&46&63)] −[■8(23×1&23×2&23×3@23×3&23×(−2)&23×1@23×4&23×(2)&23×1)] − [■8(1×40&0×40&0×40@0×40&1×40&0×40@0×40&0×40&1×40)] = [■8(63&46&69@69&−6&23@92&46&63)] ⤶7− [■8(23&46&69@69&−46&23@92&46&23)] ⤶7− [■8(40&0&0@0&40&0@0&0&40)] = [■8(63−23−40&46−46+0&69−69+0@69−69+0&−6+46−40&23−23+0@92−92+0&46−46+0&63−23−40)] = [■8(𝟎&𝟎&𝟎@𝟎&𝟎&𝟎@𝟎&𝟎&𝟎)] = O Hence proved.

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