# How do you find the determinant of ((12, 5, -2), ( -3, 0, 1), (-5, 4, 2))?

Aug 20, 2016

$| \left(12 , 5 , - 2\right) , \left(- 3 , 0 , 1\right) , \left(- 5 , 4 , 2\right) | = \textcolor{g r e e n}{- 19}$

#### Explanation:

Method 1 (what your teacher probably wanted)
$| \left(\textcolor{b l u e}{12} , \textcolor{b l u e}{5} , \textcolor{b l u e}{- 2}\right) , \left(- 3 , 0 , 1\right) , \left(- 5 , 4 , 2\right) |$

=color(red)(+)color(blue)(12)|(color(purple)(0),color(green)(1)),(color(green)(4),color(purple)(2))|color(red)(-)color(blue)(5)|(color(purple)(-3),color(green)(1)),(color(green)(-5),color(purple)(2))|color(red)(+)color(blue)(""(-2))|(color(purple)(-3),color(green)(0)),(color(green)(-5),color(purple)(4))|

$= \textcolor{b l u e}{12} \left(\textcolor{p u r p \le}{0 \times 2} - \textcolor{g r e e n}{4 \times 1}\right)$
$\textcolor{w h i t e}{\text{X}} - \textcolor{b l u e}{5} \left(\textcolor{p u r p \le}{\left(- 3\right) \times 2} - \textcolor{g r e e n}{\left(- 5\right) \times 1}\right)$
$\textcolor{w h i t e}{\text{X}} \textcolor{b l u e}{- 2} \left(\textcolor{p u r p \le}{\left(- 3\right) \times 4} - \textcolor{g r e e n}{\left(- 5\right) \times 0}\right)$

$= \textcolor{b l u e}{12} \left(\textcolor{\mathmr{and} a n \ge}{- 4}\right) - \textcolor{b l u e}{5} \left(\textcolor{\mathmr{and} a n \ge}{- 1}\right) \textcolor{b l u e}{- 2} \left(\textcolor{\mathmr{and} a n \ge}{- 12}\right)$

$= - 48 + 5 + 24$

$= - 19$

You might have been expected to exchange a pair of columns (or a pair of rows) to reduce the calculations a bit. Just remember:
$\textcolor{w h i t e}{\text{XXX}} | \left(12 , 5 , - 2\right) , \left(- 3 , 0 , 1\right) , \left(- 5 , 4 , 2\right) | = \textcolor{red}{-} | \left(- 3 , 0 , 1\right) , \left(12 , 5 , 2\right) , \left(- 5 , 4 , 2\right) |$
(exchanging rows or columns negates the value of the determinant).

Method 2 (how you would do this if you needed it in real life)
Enter the matrix into a spreadsheet and use a builtin function: