Check sibling questions

 


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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.

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise

About the Author

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 and Computer Science at Teachoo