# How do you find the determinant of ((2, 5, -3, -1), (3, 0, 1, -3), (-4, 5, -7, 8), (4, 10, -4, 1))?

Apr 30, 2016

$180$

#### Explanation:

$\left[\begin{matrix}\textcolor{t e a l}{2} & \textcolor{t e a l}{5} & \textcolor{t e a l}{- 3} & \textcolor{t e a l}{- 1} \\ 3 & 0 & 1 & - 3 \\ \textcolor{m a \ge n t a}{- 4} & \textcolor{m a \ge n t a}{5} & \textcolor{m a \ge n t a}{- 7} & \textcolor{m a \ge n t a}{8} \\ \textcolor{t u r q u o i s e}{4} & \textcolor{t u r q u o i s}{10} & \textcolor{t u r q u o i s e}{- 4} & \textcolor{t u r q u o i s e}{1}\end{matrix}\right] =$
$\to$ row3 $-$ row1 $\to$ row3
$\to$ row4 $- 2 \cdot$ row1 $\to$ row4

$= \left[\begin{matrix}\textcolor{\mathmr{and} a n \ge}{2} & \textcolor{b l u e}{5} & - 3 & - 1 \\ \textcolor{\mathmr{and} a n \ge}{3} & \textcolor{b l u e}{0} & 1 & - 3 \\ \textcolor{\mathmr{and} a n \ge}{- 6} & \textcolor{b l u e}{0} & - 4 & 9 \\ \textcolor{\mathmr{and} a n \ge}{0} & \textcolor{b l u e}{0} & 2 & 3\end{matrix}\right]$
$\to$ changing column1 with colum2

$= - \left[\begin{matrix}\textcolor{b l u e}{5} & 2 & - 3 & - 1 \\ 0 & 3 & 1 & - 3 \\ 0 & - 6 & - 4 & 9 \\ 0 & 0 & 2 & 3\end{matrix}\right]$

$= - 5. {\left(- 1\right)}^{1 + 1} \cdot \left[\begin{matrix}\textcolor{v i o \le t}{3} & 1 & \textcolor{c r i m s o n}{- 3} \\ \textcolor{v i o \le t}{- 6} & - 4 & \textcolor{c r i m s o n}{9} \\ \textcolor{v i o \le t}{0} & 2 & \textcolor{c r i m s o n}{3}\end{matrix}\right]$

$= - 5 \cdot 3 \cdot 3 \cdot \left[\begin{matrix}1 & 1 & - 1 \\ - 2 & - 4 & 3 \\ 0 & 2 & 1\end{matrix}\right]$

$= - 45 \cdot \left[\cancel{- 4} + \cancel{4} - \left(6 - 2\right)\right] = - 45 \cdot \left[- 4\right] = 180$