Question

19. Let \( A=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0\end{array}\right) \). Then \( A^{2025}-A^{2020} \) is equal to [JEE (Main)-2021] (1) \( A^{6}-A \) (2) \( A^{5} \) (3) \( A^{5}-A \) (4) \( A^{6} \)

Answer 1

Correct option is (A) A⁶ - A

\(|A - \lambda| = \begin{vmatrix} 1-\lambda&0&0\\0&1-\lambda &1\\1&0&-\lambda\end{vmatrix} = 0\)

⇒ λ³ − 2λ² + λ = 0

A³ − 2A² + A = 0 

⇒ A² − A = A³ − A² = A⁴ − A³ = A⁵ − A⁴ = A⁶ − A⁵ = A⁷ − A⁶

So, A⁷ − A² = A⁶ − A

⇒ A⁸ − A³ = A⁷ − A² = A⁶ − A

And so on.

∴ A²⁰²⁵ − A²⁰²⁰ = A⁶ − A