Question 3 - Chapter 3 Class 12 Matrices (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 3 Class 12 Matrices
Ex 3.1, 9 (MCQ) Important
Example 18 Important
Example 19
Ex 3.2, 7 (i)
Ex 3.2, 12 Important
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 20 Important
Example 22 Important
Ex 3.3, 4 Important
Ex 3.3, 10 (i) Important
Ex 3.3, 12 (MCQ)
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Example 25
Question 3 Important Deleted for CBSE Board 2024 Exams You are here
Misc 6 Important
Misc 8 Important
Misc 9 (MCQ)
Chapter 3 Class 12 Matrices
Last updated at April 16, 2024 by Teachoo
Question 3 If A = [■8(3&−4@1&−1)] , then prove An = [■8(1+2n&−4n@n&1−2n)] where n is any positive integer We shall prove the result by using mathematical induction. Step 1: P(n): If A= [■8(3&−4@1&−1)] , then An = [■8(1+2n&−4n@n&1−2n)] , n ∈ N Step 2: Prove for n = 1 For n = 1 L.H.S = A1 = A = [■8(3&−4@1&−1)] R.H.S = [■8(1+2(1)&−4(1)@1&1−2(1))]=[■8(1+2&−4@1&1−2)]" = " [■8(3&−4@1&−1)] L.H.S = R.H.S ∴ P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k+1) is true Assume that P (k) is true P(k) : If A= [■8(3&−4@1&−1)] , then Ak = [■8(1+2k&−4k@k&1−2k)] We will have to prove that P( k +1) is true P(k + 1) : If A= [■8(3&−4@1&−1)] , then Ak+1 = [■8(1+2(k+1)&−4(k+1)@(k+1)&1−2(k+1))] Taking L.H.S Ak+1 = Ak . A = [■8(1+2k&−4k@k&1−2k)] [■8(3&−4@1&−1)] = [■8((1+2k)3−4k(1)&(1+2k)(−4)−4k(−1)@k(3)+(1−2k)1&k(−4)+(1−2k)(−1))] = [■8(3+6k−4k&−4−8k+4k@3k+1−2k&−4k−1+2k)] = [■8(3+2k&−4−4k@1+k&−1−2k)] = [■8(1+2(k+1)&−4(k+1)@1+k&1−2(k+1))] = R.H.S Thus P (k + 1) is true ∴ By the principal of mathematical induction , P(n) is true for n ∈ N Hence, if A= [■8(3&−4@1&−1)] , then An = [■8(1+2n&−4n@n&1−2n)] , n ∈ N