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

Jun 3, 2016

It is $- 30$.

#### Explanation:

For a $2 \setminus \times 2$ matrix the determinant is

$\det \left[\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\right] = a d - b c$

In a $3 \setminus \times 3$ matrix you have to calculate the determinant of three sub-matrices $2 \setminus \times 2$.

We begin selecting the first element

$\left(\begin{matrix}\setminus \textcolor{red}{a} & b & c \\ d & e & f \\ g & h & i\end{matrix}\right)$

then we identify the sub-matrix that is not in the raw or the column of that element

$\left(\begin{matrix}\setminus \textcolor{red}{a} & b & c \\ d & \setminus \textcolor{c y a n}{e} & \setminus \textcolor{c y a n}{f} \\ g & \setminus \textcolor{c y a n}{h} & \setminus \textcolor{c y a n}{i}\end{matrix}\right)$

and finally we multiply the element selected (in our case $a$) by the determinant of the $2 \setminus \times 2$ matrix

$a \left(e i - f h\right)$.

We repeat this process for the elements of the first raw alternating the signs

$\left(\begin{matrix}a & \setminus \textcolor{red}{b} & c \\ \setminus \textcolor{c y a n}{d} & e & \setminus \textcolor{c y a n}{f} \\ \setminus \textcolor{c y a n}{g} & h & \setminus \textcolor{c y a n}{i}\end{matrix}\right)$

$- b \left(\mathrm{di} - f g\right)$

$\left(\begin{matrix}a & b & \setminus \textcolor{red}{c} \\ \setminus \textcolor{c y a n}{d} & \setminus \textcolor{c y a n}{e} & f \\ \setminus \textcolor{c y a n}{g} & \setminus \textcolor{c y a n}{h} & i\end{matrix}\right)$

$c \left(\mathrm{dh} - g e\right)$.

The determinant is then the sum of these three terms:

$a \left(e i - f h\right) - b \left(\mathrm{di} - f g\right) + c \left(\mathrm{dh} - g e\right)$

We are now ready to calculate our determinant:

$\det \left(\begin{matrix}5 & 2 & 4 \\ 2 & - 2 & - 3 \\ - 3 & 2 & 5\end{matrix}\right) =$

$5 \cdot \left(- 2 \cdot 5 - \left(- 3\right) \cdot 2\right)$

$- 2 \cdot \left(2 \cdot 5 - \left(- 3\right) \cdot \left(- 3\right)\right)$

$+ 4 \cdot \left(2 \cdot 2 - \left(- 2\right) \cdot \left(- 3\right)\right)$

$= - 20 - 2 - 8 = - 30$.