If A,B,C are three square matrices of same order satisfying A2 = A–1 and let B=A2n and C=A2n−2(n∈N) then det (B-C) =

B=A2n=A2⋅2n−1=A−12n−1=A−12⋅2n−2=A−122n−2=A2−12n−2=A2n−2=C⇒B−C=0⇒|B−C|=0

If A,B,C are three square matrices of same order satisfying A2 = A–1 and let B=A2n and C=A2n−2(n∈N) then det (B-C) =

B=A2n=A2⋅2n−1=A−12n−1=A−12⋅2n−2=A−122n−2=A2−12n−2=A2n−2=C⇒B−C=0⇒|B−C|=0

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If A,B,C are three square matrices of same order satisfying A² = A^–1 and let $B = A^{2^{n}}$ and $C = A^{2^{n - 2}} (n \in N)$ then det (B-C) =

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

-1

$(- 1)^{n}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} B = A^{2^{n}} = A^{2 \cdot 2^{n - 1}} = {(A^{- 1})}^{2^{n - 1}} = {(A^{- 1})}^{2 \cdot 2^{n - 2}} \\ = {({(A^{- 1})}^{2})}^{2^{n - 2}} = {({(A^{2})}^{- 1})}^{2^{n - 2}} = A^{2^{n - 2}} = C \\ \Rightarrow B - C = 0 \Rightarrow | B - C | = 0 \end{array}$