I have the following matrix of order four for which I have calculated the determinant using Laplace's method.
$$
\begin{bmatrix}
2 & 1 & 3 & 1 \\
4 & 3 & 1 & 4 \\
-1 & 5 & -2 & 1 \\
1 & 3 & -2 & -1 \\
\end{bmatrix}
$$
Finding the determinant gives me $-726$. Now if I check the result at Wolfram Alpha, it says the result is $-180$ (Because there are no zeros in the matrix, expand with respect to row one) so it uses only the first row to calculate the determinant of the matrix.
My question is: Why it uses only the first row to find the determinant?
Best Answer
The Laplace development can be performed with respect to any row or column. Let's see when developing with respect to the first row: \begin{align} \det\begin{bmatrix} 2 & 1 & 3 & 1 \\ 4 & 3 & 1 & 4 \\ -1 & 5 & -2 & 1 \\ 1 & 3 & -2 & -1 \end{bmatrix}={}& (-1)^{1+1}\cdot 2 \cdot \det\begin{bmatrix} 3 & 1 & 4 \\ 5 & -2 & 1 \\ 3 & -2 & -1 \end{bmatrix}+{}\\ &(-1)^{1+2}\cdot 1 \cdot \det\begin{bmatrix} 4 & 1 & 4 \\ -1 & -2 & 1 \\ 1 & -2 & -1 \end{bmatrix}+{}\\ &(-1)^{1+3}\cdot 3\cdot \det\begin{bmatrix} 4 & 3 & 4 \\ -1 & 5 & 1 \\ 1 & 3 & -1 \end{bmatrix}+{}\\ &(-1)^{1+4}\cdot1\cdot \det\begin{bmatrix} 4 & 3 & 1 \\ -1 & 5 & -2 \\ 1 & 3 & -2 \end{bmatrix} \end{align} Now you can go on by computing the determinants of the four $3\times3$ matrices with the same (or another) method. The final result is indeed $-180$.
When one row or column has many zeroes it's convenient to use that one, but any row or column can be used.