How do I find the determinant of this matrix using expansion by cofactors?
$$\begin{bmatrix} 2& 7& −1& 4\\ 0& −5& 8& 11\\ 0& 0& 3& −13\\ 0& 0& 0& 1 \end{bmatrix}$$
I understood how to do this with a smaller matrix but I'm not sure how to do it for a larger matrix like this.
Thank you
To minimize calculations, you want to expand the determinant along a row/column that has as many zeros as possible. For example, expanding along the first column, we have
$$ \det \begin{pmatrix} 2 & 7 & -1 & 4 \\ 0 & -5 & 8 & 11 \\ 0 & 0 & 3 & -13 \\ 0 & 0 & 0 & 1 \end{pmatrix} = 2 \cdot \det \begin{pmatrix} -5 & 8 & 11 \\ 0 & 3 & -13 \\ 0 & 0 & 1\end{pmatrix} - 0\cdot \det \begin{pmatrix} 7 & -1 & 4 \\ 0 & 3 & -13 \\ 0 & 0 & 1 \end{pmatrix} \\ + 0 \cdot \det \begin{pmatrix} 7 & -1 & 4 \\ -5 & 8 & 11 \\ 0 & 0 & 1 \end{pmatrix} - 0 \cdot \det \begin{pmatrix} 7 & -1 & 4 \\ -5 & 8 & 11 \\ 0 & 3 & -13 \end{pmatrix} \\ = 2 \cdot \det \begin{pmatrix} -5 & 8 & 11 \\ 0 & 3 & -13 \\ 0 & 0 & 1\end{pmatrix}. $$
Doing the same for the resulting $3 \times 3$ matrix (that is, expanding along the first column), we get
$$ \det \begin{pmatrix} 2 & 7 & -1 & 4 \\ 0 & -5 & 8 & 11 \\ 0 & 0 & 3 & -13 \\ 0 & 0 & 0 & 1 \end{pmatrix} = 2 \cdot \det \begin{pmatrix} -5 & 8 & 11 \\ 0 & 3 & -13 \\ 0 & 0 & 1\end{pmatrix} = 2 \cdot (-5) \cdot \det \begin{pmatrix} 3 & -13 \\ 0 & 1 \end{pmatrix} $$
and so on.
To make sure you understand the technique, try doing the calculation by expanding along the last row every time and see that you get the same answer ($-30$). Then to a random row/column expansion and verify that you still get the same answer.
