Let \(A = \begin{pmatrix} 2026 & 2025 & 2024 \\ 2025 & 2024 & 2023 \\ 2024 & 2023 & 2023 \end{pmatrix}\) be a matrix. Then, \(\det(A)\) is equal to _______
Show Hint
- -1
- 0
- 1
- -4051
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
Find the determinant of the following \(3 \times 3\) matrix:
\[ A = \begin{pmatrix} 2026 & 2025 & 2024 \\ 2025 & 2024 & 2023 \\ 2024 & 2023 & 2023 \end{pmatrix} \]
Since the entries are large, direct expansion is lengthy. We will simplify the matrix using row operations.
Step 2: Key Formula or Approach:
We use the property of determinants:
Adding or subtracting a multiple of one row from another row does not change the value of the determinant.
\[ \det(A) = \det(A') \]
Step 3: Detailed Explanation:
Apply the row operation: \[ R_1 \rightarrow R_1 - R_2 \] The first row becomes: \[ R_1 = (2026-2025,\;2025-2024,\;2024-2023) \] \[ R_1 = (1,\;1,\;1) \]
Apply the row operation: \[ R_2 \rightarrow R_2 - R_3 \] The second row becomes: \[ R_2 = (2025-2024,\;2024-2023,\;2023-2023) \] \[ R_2 = (1,\;1,\;0) \]
Therefore, the transformed matrix is: \[ A' = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 2024 & 2023 & 2023 \end{pmatrix} \]
Expanding the determinant along the first row: \[ \det(A') = 1 \begin{vmatrix} 1 & 0 \\ 2023 & 2023 \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \\ 2024 & 2023 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 2024 & 2023 \end{vmatrix} \]
Now calculate the minors: \[ \det(A') = 1(1 \times 2023 - 0 \times 2023) - 1(1 \times 2023 - 0 \times 2024) + 1(1 \times 2023 - 1 \times 2024) \] \[ \det(A') = 2023 - 2023 + (2023-2024) \] \[ \det(A') = -1 \]
Step 4: Final Answer:
The determinant of the given matrix is: \[ \boxed{-1} \]
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