JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(\mathrm{A}^{6}\)
Answer & Solution
Correct Answer
(A) \(A^{6}-A\)
Step-by-step Solution
Detailed explanation
\(A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{array}\right] \Rightarrow A^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{array}\right]\)…
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