Suppose that \begin{equation*} T= \begin{bmatrix} 2 & 3 & -1 & 3 & 3\\ 0 & -1 & 5 & 2 & -1\\ 0 & 0 & 3 & 9 & 2\\ 0 & 0 & 0 & -1 & 3\\ 0 & 0 & 0 & 0 & 5 \end{bmatrix} \end{equation*}
We will compute the determinant of this $5\times 5$ matrix by consistently expanding about the first column for each submatrix that arises and does not have a zero entry multiplying it. \begin{align*} \detname{T}&= \begin{vmatrix} 2 & 3 & -1 & 3 & 3\\ 0 & -1 & 5 & 2 & -1\\ 0 & 0 & 3 & 9 & 2\\ 0 & 0 & 0 & -1 & 3\\ 0 & 0 & 0 & 0 & 5 \end{vmatrix}\\ &=2(-1)^{1+1} \begin{vmatrix} -1 & 5 & 2 & -1\\ 0 & 3 & 9 & 2\\ 0 & 0 & -1 & 3\\ 0 & 0 & 0 & 5 \end{vmatrix}\\ &=2(-1)(-1)^{1+1} \begin{vmatrix} 3 & 9 & 2\\ 0 & -1 & 3\\ 0 & 0 & 5 \end{vmatrix}\\ &=2(-1)(3)(-1)^{1+1} \begin{vmatrix} -1 & 3\\ 0 & 5 \end{vmatrix}\\ &=2(-1)(3)(-1)(-1)^{1+1} \begin{vmatrix} 5 \end{vmatrix}\\ &=2(-1)(3)(-1)(5)=30 \end{align*}