# How do you find the value of the determinant |(6, 3, -1), (0, 3, 3), (-9, 0, 0)|?

Jan 7, 2017

$\left\mid \begin{matrix}6 & 3 & - 1 \\ 0 & 3 & 3 \\ - 9 & 0 & 0\end{matrix} \right\mid = - 108$

#### Explanation:

I will use a couple of properties of the determinant:

• The determinant is unaltered when a multiple of one row is added to (or subtracted from) another row.

• The determinant of an upper ot lower triangular matrix is the product of the main diagonal.

Given:

$\left\mid \begin{matrix}6 & 3 & - 1 \\ 0 & 3 & 3 \\ - 9 & 0 & 0\end{matrix} \right\mid$

Subtract $\text{row2}$ from $\text{row1}$ to get:

$\left\mid \begin{matrix}6 & 0 & - 4 \\ 0 & 3 & 3 \\ - 9 & 0 & 0\end{matrix} \right\mid$

Add $\frac{3}{2} \times \text{row1}$ to $\text{row3}$ to get:

$\left\mid \begin{matrix}6 & 0 & - 4 \\ 0 & 3 & 3 \\ 0 & 0 & - 6\end{matrix} \right\mid$

Since this is an upper triangular matrix, we can just multiply the diagonal to get the determinant:

$\left\mid \begin{matrix}\textcolor{b l u e}{6} & 0 & - 4 \\ 0 & \textcolor{b l u e}{3} & 3 \\ 0 & 0 & \textcolor{b l u e}{- 6}\end{matrix} \right\mid = 6 \cdot 3 \cdot \left(- 6\right) = - 108$