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

May 2, 2016

$- 47$

#### Explanation:

$\left[\begin{matrix}\textcolor{t e a l}{2} & 0 & - 1 & \textcolor{\mathmr{and} a n \ge}{3} \\ \textcolor{t e a l}{- 1} & 1 & 0 & \textcolor{\mathmr{and} a n \ge}{2} \\ \textcolor{t e a l}{0} & 3 & 2 & \textcolor{\mathmr{and} a n \ge}{4} \\ \textcolor{t e a l}{4} & 1 & - 1 & \textcolor{\mathmr{and} a n \ge}{0}\end{matrix}\right] =$
$\to$ column4 $+ 2 \cdot$ column1 $\to$ column4

=$\left[\begin{matrix}\textcolor{t e a l}{2} & \textcolor{\mathmr{and} a n \ge}{0} & - 1 & 7 \\ \textcolor{t e a l}{- 1} & \textcolor{\mathmr{and} a n \ge}{1} & 0 & 0 \\ \textcolor{t e a l}{0} & \textcolor{\mathmr{and} a n \ge}{3} & 2 & 4 \\ \textcolor{t e a l}{4} & \textcolor{\mathmr{and} a n \ge}{1} & - 1 & 8\end{matrix}\right]$
$\to$ column1 $+$ column2 $\to$ column1

$= \left[\begin{matrix}2 & \textcolor{g o l d}{0} & - 1 & 7 \\ \textcolor{g o l d}{0} & \textcolor{b l u e}{1} & \textcolor{g o l d}{0} & \textcolor{g o l d}{0} \\ 3 & \textcolor{g o l d}{3} & 2 & 4 \\ 5 & \textcolor{g o l d}{1} & - 1 & 8\end{matrix}\right]$

$= 1 \cdot {\left(- 1\right)}^{2 + 2} \left[\begin{matrix}2 & - 1 & 7 \\ 3 & 2 & 4 \\ 5 & - 1 & 8\end{matrix}\right]$
$= 32 - 20 - 21 - \left(70 - 8 - 24\right) = - 9 - 38 = - 47$