Question:
The determinant of the matrix
\[
\begin{pmatrix}
2021 & 2020 & 2020 & 2020 \\
2021 & 2021 & 2020 & 2020 \\
2021 & 2021 & 2021 & 2020 \\
2021 & 2021 & 2021 & 2021
\end{pmatrix}
\]
is _________.
The determinant of the matrix
\[
\begin{pmatrix}
2021 & 2020 & 2020 & 2020 \\
2021 & 2021 & 2020 & 2020 \\
2021 & 2021 & 2021 & 2020 \\
2021 & 2021 & 2021 & 2021
\end{pmatrix}
\]
is _________.
Show Hint
When rows differ by a constant, subtracting adjacent rows simplifies the matrix to a triangular form,
making determinant evaluation straightforward.
Updated On: Dec 6, 2025
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Correct Answer: 2021
Solution and Explanation
Step 1: Simplify using row operations.
Start with \[ \begin{pmatrix} 2021 & 2020 & 2020 & 2020 \\ 2021 & 2021 & 2020 & 2020 \\ 2021 & 2021 & 2021 & 2020 \\ 2021 & 2021 & 2021 & 2021 \end{pmatrix}. \] Perform successive row differences: \[ R_2 \to R_2 - R_1,\qquad R_3 \to R_3 - R_2 \text{ (original)},\qquad R_4 \to R_4 - R_3 \text{ (original)}. \] This gives the upper-triangular matrix \[ \begin{pmatrix} 2021 & 2020 & 2020 & 2020 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \]
Step 2: Compute determinant.
For an upper-triangular matrix the determinant is the product of diagonal entries: \[ \det = 2021 \times 1 \times 1 \times 1 = 2021. \]
Step 3: Conclusion.
Hence, the determinant is \(\boxed{2021}.\)
Start with \[ \begin{pmatrix} 2021 & 2020 & 2020 & 2020 \\ 2021 & 2021 & 2020 & 2020 \\ 2021 & 2021 & 2021 & 2020 \\ 2021 & 2021 & 2021 & 2021 \end{pmatrix}. \] Perform successive row differences: \[ R_2 \to R_2 - R_1,\qquad R_3 \to R_3 - R_2 \text{ (original)},\qquad R_4 \to R_4 - R_3 \text{ (original)}. \] This gives the upper-triangular matrix \[ \begin{pmatrix} 2021 & 2020 & 2020 & 2020 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \]
Step 2: Compute determinant.
For an upper-triangular matrix the determinant is the product of diagonal entries: \[ \det = 2021 \times 1 \times 1 \times 1 = 2021. \]
Step 3: Conclusion.
Hence, the determinant is \(\boxed{2021}.\)
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