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

Feb 9, 2017

$| \left(2 , - 1 , 3\right) , \left(3 , 0 , - 2\right) , \left(1 , - 3 , 0\right) | = - 37$

#### Explanation:

If you take the determinant of a $3 \times 3$ matrix
$| \left(a , b , c\right) , \left(f , g , h\right) , \left(x , y , z\right) |$
you take each element in the first row and multiply it by the determinants of $2 \times 2$ submatrices which are composed by the elements in the bottom two rows of each column other than the column the particular first row element is from

$| \left(a , b , c\right) , \left(f , g , h\right) , \left(x , y , z\right) | = a | \left(g , h\right) , \left(y , z\right) | - b | \left(f , h\right) , \left(x , z\right) | + c | \left(f , g\right) , \left(x , y\right) |$

Notice the second term is negatve. The signs alternate.

Then we take the determinants of the submatrices.

$| \left(a , b , c\right) , \left(f , g , h\right) , \left(x , y , z\right) | = a \left(g z - h y\right) - b \left(f z - h x\right) + c \left(f y - g x\right)$

In this case our matrix is
$| \left(2 , - 1 , 3\right) , \left(3 , 0 , - 2\right) , \left(1 , - 3 , 0\right) |$

Then

$| \left(2 , - 1 , 3\right) , \left(3 , 0 , - 2\right) , \left(1 , - 3 , 0\right) | = 2 | \left(0 , - 2\right) , \left(- 3 , 0\right) | - \left(- 1\right) | \left(3 , - 2\right) , \left(1 , 0\right) | + 3 | \left(3 , 0\right) , \left(1 , - 3\right) |$

$= 2 \left(0 - 6\right) + \left(0 - \left(- 2\right)\right) + 3 \left(- 9 - 0\right)$

$= 2 \left(- 6\right) + 2 + 3 \left(- 9\right) = - 12 + 2 - 27$

$= - 12 - 27 + 2 = - 39 + 2 = - 37$