# How do you find the determinant of ((2, 5, -3, -1), (3, 0, 1, -3), (-6, 0, -4, 9), ( 4, 10, -4, -1))?

May 3, 2016

$120$

#### Explanation:

$\left(\begin{matrix}\textcolor{t e a l}{2} & \textcolor{t e a l}{5} & \textcolor{t e a l}{- 3} & \textcolor{t e a l}{- 1} \\ 3 & 0 & 1 & - 3 \\ - 6 & 0 & - 4 & 9 \\ \textcolor{t u r q u o i s e}{4} & \textcolor{t u r q u o i s e}{10} & \textcolor{t u r q u o i s e}{- 4} & \textcolor{t u r q u o i s e}{- 1}\end{matrix}\right) =$
$\to$ row4 $- 2 \cdot$ row1 $\to$ row4

$= \left(\begin{matrix}\textcolor{g o l d}{2} & \textcolor{b l u e}{5} & \textcolor{g o l d}{- 3} & \textcolor{g o l d}{- 1} \\ 3 & \textcolor{g o l d}{0} & 1 & - 3 \\ - 6 & \textcolor{g o l d}{0} & - 4 & 9 \\ 0 & \textcolor{g o l d}{0} & 2 & 1\end{matrix}\right)$

$= 5 \cdot {\left(- 1\right)}^{1 + 2} \cdot \left(\begin{matrix}3 & 1 & - 3 \\ - 6 & - 4 & 9 \\ 0 & 2 & 1\end{matrix}\right)$

$= - 5 \left[- 12 + 36 - \left(54 - 6\right)\right]$
$= - 5 \left[24 - 48\right] = 5 \cdot 24 = 120$