If A is a triangular matrix, det \(A\) is the product of the entries on the main diagonal of A: $$\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ 0& a_{22}&a_{23}\\0& 0&a_{33}\end{array}\right|=a_{11}a_{22}a_{33}$$
Proof
We prove when \(n=3\) .
If we expand the determinant of LHS down the 1st column, we have
$$\begin{eqnarray*}\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ 0& a_{22}&a_{23}\\0& 0&a_{33}\end{array}\right|&=&(-1)^{1+1}a_{11}\left|\begin{array}{cc}a_{22}& a_{23}\\0& a_{33}\end{array}\right|+(-1)^{2+1}0\left|\begin{array}{cc}a_{12}& a_{13}\\ 0& a_{33}\end{array}\right|+(-1)^{3+1}0\left|\begin{array}{cc}a_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|\\ && \\&=&a_{11}\left|\begin{array}{cc}a_{22}& a_{23}\\ 0& a_{33}\end{array}\right|\\ &&\\&=&a_{11}a_{22}a_{33}\end{eqnarray*}$$
In general case of n, we can prove by induction.