# Misc 5 - Chapter 3 Class 12 Matrices

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5 You are here

Misc 6 Important

Misc 7 Important

Misc 8 Important

Misc 9 (MCQ)

Misc 10 (MCQ) Important

Misc 11 (MCQ) Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Deleted for CBSE Board 2024 Exams

Chapter 3 Class 12 Matrices

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 5 If A = [■8(3&1@−1&2)] , show that A2 – 5A + 7I = O First calculating A2 A2 = A.A A2 = [■8(𝟑&𝟏@−𝟏&𝟐)] [■8(𝟑&𝟏@−𝟏&𝟐)] = [■8(3(3)+1(−1)&3(1)+1(2)@−1(3)+2(−1)&−1(1)+2(2))] = [■8(9−1&3+2@−3−2&−1+4)] = [■8(𝟖&𝟓@−𝟓&𝟑)] Now calculating A2 – 5A – 7I = [■8(𝟖&𝟓@−𝟓&𝟑)] – 5 [■8(𝟑&𝟏@−𝟏&𝟐)] + 7 [■8(𝟏&𝟎@𝟎&𝟏)] = [■8(8&5@−5&3)] – [■8(5(3)&5(1)@5(−1)&5(2))] + [■8(7(1)&7(0)@7(0)&7(1))] = [■8(8&5@−5&3)] – [■8(15&5@−5&10)] + [■8(7&0@0&7)] = [■8(8−15+7&5−5+0@−5−(−5)&3−10+7)] = [■8(𝟎&𝟎@𝟎&𝟎)] = O = R.H.S. Since L.H.S = R.H.S Hence proved