3 and 4 .Determinants and Matrices
medium

Let $\mathrm{A}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0\end{array}\right) .$ Then $\mathrm{A}^{2025}-\mathrm{A}^{2020}$ is equal to :

A

$A^{6}-A$

B

$\mathrm{A}^{5}$

C

$\mathrm{A}^{5}-\mathrm{A}$

D

$\mathrm{A}^{6}$

(JEE MAIN-2021)

Solution

$A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{array}\right] \Rightarrow A^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{array}\right]$

$A^{3}=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0\end{array}\right] \Rightarrow A^{4}=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 1 \\ 1 & 0 & 0\end{array}\right]$

$\mathrm{A}^{\mathrm{n}}=\left[\begin{array}{ccc}1 & 0 & 0 \\ \mathrm{n}-1 & 1 & 1 \\ 1 & 0 & 0\end{array}\right]$

$\mathrm{A}^{2025}-\mathrm{A}^{2020}=\left[\begin{array}{lll}0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

$A^{6}-A=\left[\begin{array}{lll}0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

Standard 12
Mathematics

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