# How do you find the determinant of ((5, 7, -1, 1, 0, 0), (2, 2, 1, 3, -1, 0), (-3, 4, -1, -2, 1, 2), (2, 9, -3, 0, 0, 0), (0, 1, 4, 0, 0, 0), (0, 0, 1, 0, 0, 0))?

Apr 28, 2016

$4$

#### Explanation:

$\left[\begin{matrix}\textcolor{t e a l}{5} & \textcolor{t e a l}{7} & \textcolor{t e a l}{- 1} & \textcolor{t e a l}{1} & \textcolor{t e a l}{0} & \textcolor{t e a l}{0} \\ 2 & 2 & 1 & 3 & - 1 & 0 \\ \textcolor{b l u e}{- 3} & \textcolor{b l u e}{4} & \textcolor{b l u e}{- 1} & \textcolor{b l u e}{- 2} & \textcolor{b l u e}{1} & \textcolor{b l u e}{2} \\ 2 & 9 & - 3 & 0 & 0 & 0 \\ 0 & 1 & 4 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\end{matrix}\right]$
$\to$ changing row1 with row3

$= - \left[\begin{matrix}\textcolor{m a \ge n t a}{- 3} & 4 & \textcolor{\mathmr{and} a n \ge}{- 1} & - 2 & 1 & 2 \\ \textcolor{m a \ge n t a}{2} & 2 & \textcolor{\mathmr{and} a n \ge}{1} & 3 & - 1 & 0 \\ \textcolor{m a \ge n t a}{5} & 7 & \textcolor{\mathmr{and} a n \ge}{- 1} & 1 & 0 & 0 \\ \textcolor{m a \ge n t a}{2} & 9 & \textcolor{\mathmr{and} a n \ge}{- 3} & 0 & 0 & 0 \\ \textcolor{m a \ge n t a}{0} & 1 & \textcolor{\mathmr{and} a n \ge}{4} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\end{matrix}\right]$
$\to$ changing column1 with column3

$= \left[\begin{matrix}- 1 & 4 & - 3 & - 2 & 1 & 2 \\ 1 & 2 & 2 & 3 & - 1 & 0 \\ - 1 & 7 & 5 & 1 & 0 & 0 \\ - 3 & 9 & 2 & 0 & 0 & 0 \\ 4 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0\end{matrix}\right]$

$= - \left[1 \cdot 1 \cdot 2 \cdot 1 \cdot \left(- 1\right) \cdot 2\right] = 4$

A brief explanation of the MINUS sign before the multiplication in the last line:
Considering the last determinant, how many times does one have to change columns to arrive at the shape

$\left[\begin{matrix}a & b & c & d & e & f \\ 0 & g & h & i & j & k \\ 0 & 0 & l & m & n & o \\ 0 & 0 & 0 & p & q & r \\ 0 & 0 & 0 & 0 & s & t \\ 0 & 0 & 0 & 0 & 0 & u\end{matrix}\right]$?

Answer: 3 times (column1 with column6, column2 with column5 and column3 with column4)
Each time the columns are changed with each other the sign is changed, so:

${\left(- 1\right)}^{3} = - 1$ (that's why the multiplication is preceded by a minus sign)