# How do you find the determinant of ((1, 1, 1, 3), (0, 3, 1, 1), (0, 0, 2, 2), (-1, -1, -1, 2))?

Mar 3, 2016

$\left\mid \begin{matrix}1 & 1 & 1 & 3 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 2 & 2 \\ - 1 & - 1 & - 1 & 2\end{matrix} \right\mid = 30$

#### Explanation:

The determinant is unchanged by adding any multiple of one row to another, so first add row $1$ to row $4$ to find:

$\left\mid \begin{matrix}1 & 1 & 1 & 3 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 2 & 2 \\ - 1 & - 1 & - 1 & 2\end{matrix} \right\mid = \left\mid \begin{matrix}1 & 1 & 1 & 3 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 5\end{matrix} \right\mid$

The determinant of an upper triangular matrix is just the product of the diagonal, so:

$\left\mid \begin{matrix}1 & 1 & 1 & 3 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 5\end{matrix} \right\mid = 1 \cdot 3 \cdot 2 \cdot 5 = 30$