# How do you find the determinant of ((14, -13, 0), (3, 8, -1), (-10, -2, 5))?

Mar 4, 2016

$\det = 597$

#### Explanation:

There are two ways of evaluating a third-order determinant, but the most easiest is by;

$| \left({a}_{11} \quad {a}_{12} \quad {a}_{13}\right) , \left({a}_{21} \quad {a}_{22} \quad {a}_{23}\right) , \left({a}_{31} \quad {a}_{32} \quad {a}_{33}\right) | = {a}_{11} | \left({a}_{22} \quad {a}_{23}\right) , \left({a}_{32} \quad {a}_{33}\right) | - {a}_{12} | \left({a}_{21} \quad {a}_{23}\right) , \left({a}_{31} \quad {a}_{33}\right) | + {a}_{13} | \left({a}_{21} \quad {a}_{22}\right) , \left({a}_{31} \quad {a}_{32}\right) |$

$\det = {a}_{11} \left({a}_{22} {a}_{33} - {a}_{23} {a}_{32}\right) - {a}_{12} \left({a}_{21} {a}_{33} - {a}_{23} {a}_{31}\right) + {a}_{13} \left({a}_{21} {a}_{32} - {a}_{22} {a}_{31}\right)$

What needs to be remembered is that we can choose any row to break down the third-order arrangement into second-order arrangement, and it is easier when choosing a row with $0$ in it because any second-order arrangement multiplied by $0$ is $0$. but also we must consider its sign;

$| \left({a}_{11} \quad {a}_{12} \quad {a}_{13}\right) , \left({a}_{21} \quad {a}_{22} \quad {a}_{23}\right) , \left({a}_{31} \quad {a}_{32} \quad {a}_{33}\right) | = | \left(+ - +\right) , \left(- + -\right) , \left(+ - +\right) |$

When breaking down for example into ${a}_{11} | \left({a}_{22} \quad {a}_{23}\right) , \left({a}_{32} \quad {a}_{33}\right) |$, the $| \left({a}_{22} \quad {a}_{23}\right) , \left({a}_{32} \quad {a}_{33}\right) |$ is determined by eliminating the number that horizontally and vertically in line with ${a}_{11}$. By this way, ${a}_{12} , {a}_{13} , {a}_{21}$ and ${a}_{31}$ is eliminated.

So, from the question given, we break down it by using row $1$ since it has $0$ in it;

$| \left(14 , - 13 , 0\right) , \left(3 , 8 , - 1\right) , \left(- 10 , - 2 , 5\right) | =$

$= 14 | \left(8 , - 1\right) , \left(- 2 , 5\right) | - \left(- 13\right) | \left(3 , - 1\right) , \left(- 10 , 5\right) | + 0 | \left(3 , 8\right) , \left(- 10 , - 2\right) |$

$\det = 14 \left[8 \left(5\right) - \left(- 1\right) \left(- 2\right)\right] + 13 \left[3 \left(5\right) - \left(- 1\right) \left(- 10\right)\right] + 0 \left[3 \left(- 2\right) - 8 \left(- 10\right)\right]$

$\det = 14 \left(38\right) + 13 \left(5\right) + 0 \left(74\right)$

$\det = 532 + 65$

$\det = 597$