Question:
If
\[
A=
\begin{bmatrix}
2026 & 2025 & 2024\\
2025 & 2024 & 2023\\
2024 & 2023 & 2022
\end{bmatrix},
\]
then \(\left|A^{2026}-A^{2025}\right|=\)
If
\[
A=
\begin{bmatrix}
2026 & 2025 & 2024\\
2025 & 2024 & 2023\\
2024 & 2023 & 2022
\end{bmatrix},
\]
then \(\left|A^{2026}-A^{2025}\right|=\)
Show Hint
If any matrix factor has determinant zero, then the determinant of the product is also zero.
Updated On: Jul 23, 2026
- \(2026\)
- \(2025\)
- \(2024\)
- \(0\)
Show Solution
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The Correct Option is D
Solution and Explanation
Observe that
\[
A^{2026}-A^{2025}
=
A^{2025}(A-I).
\]
Taking determinant,
\[
\left|A^{2026}-A^{2025}\right|
=
|A^{2025}|\;|A-I|.
\]
Now,
\[
A-I=
\begin{bmatrix}
2025&2025&2024\\
2025&2023&2023\\
2024&2023&2021
\end{bmatrix}.
\]
Since
\[
R_1-2R_2+R_3=0,
\]
the rows are linearly dependent.
Hence,
\[
|A-I|=0.
\]
Therefore,
\[
\boxed{\left|A^{2026}-A^{2025}\right|=0.}
\]
Thus,
\[
\boxed{(D)}
\]
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