**
Question 6
**

If A = [3 1 -1 2] and I = [1 0 0 1], find k so that A
^{
2
}
= 5A + kI

Last updated at Dec. 4, 2018 by Teachoo

**
Question 6
**

If A = [3 1 -1 2] and I = [1 0 0 1], find k so that A
^{
2
}
= 5A + kI

Transcript

Question 6 If A = [■8(3&1@−1&2)] and I = [■8(1&0@0&1)], find k so that A2 = 5A + kI Finding A2 A2 = [■8(3&1@−1&2)] [■8(3&1@−1&2)] A2 = [■8(3(3)+1(−1)&3(1)+1(2)@−1(3)+2(−1)&−1(1)+2(2))] A2 = [■8(9−1&3+2@−3−2&−1+4)] A2 = [■8(8&5@−5&3)] Finding 5A 5A = 5[■8(3&1@−1&2)] 5A = [■8(5×3&5×1@5×(−1)&5×2)] 5A = [■8(15&5@−5&10)] Now, our equation is A2 = 5A + kI Putting values [■8(8&5@−5&3)] = [■8(15&5@−5&10)] + k [■8(1&0@0&1)] [■8(8&5@−5&3)] = [■8(15&5@−5&10)] + [■8(𝑘×1&𝑘×0@𝑘×0&𝑘×1)] [■8(8&5@−5&3)] = [■8(15&5@−5&10)] + [■8(𝑘&0@0&𝑘)] [■8(8&5@−5&3)] = [■8(15+𝑘&5+0@−5+0&10+𝑘)] [■8(8&5@−5&3)] = [■8(15+𝑘&5@−5&10+𝑘)] Thus, 8 = 15 + k 8 – 15 = k –7 = k k = –7

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4 (Or 1st)

Question 4 (Or 2nd)

Question 5

Question 6 You are here

Question 7

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st)

Question 10 (Or 2nd)

Question 11

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st)

Question 13 (Or 2nd)

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17

Question 18

Question 19

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23

Question 24 (Or 1st)

Question 24 (Or 2nd)

Question 25

Question 26 (Or 1st)

Question 26 (Or 2nd)

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28

Question 29

Class 12

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.