Let
\(A = \begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
0 & -1 & 2 & 3 & 4 \\
0 & 0 & 1 & 2 & 3 \\
0 & 0 & 0 & -1 & 2 \\
1 & 2 & 3 & 4 & 6
\end{bmatrix}\)?
The answer is 1.
Let \(R\) be obtained from \(A\) by adding \((-1)\) times row 1 to row 6.
Then
\(R = \begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
0 & -1 & 2 & 3 & 4 \\
0 & 0 & 1 & 2 & 3 \\
0 & 0 & 0 & -1 & 2 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}\) and
\(\det(A) = \det(R)\). But \(R\) is upper triangular.
Hence, \(\det(R)\) is given by the product of the diagonal entries,
which is \(1\cdot(-1)\cdot 1\cdot(-1) \cdot 1 = 1\). So \(\det(A) = 1\).