# How do you find the determinant of ((3, 5, 0, 6), (2, 3, 2, 0), (2, 4, 0, 7), (-3, 2, 2, 3))?

Jun 10, 2016

I found $80$

#### Explanation:

I would try to simplify a bit our original matrix to get a line or column of all zeroes but one number.

One column that seems good in this sense is the third one where we have two zeroes already!

Now, I can multiply the second line by $- 1$ and add it to the fourth line to get the other zero!

So;
the second line would become:
$- 2 , - 3 - 2 , 0$
we add this line to the third and we get:
$\left(\begin{matrix}3 & 5 & 0 & 6 \\ 2 & 3 & \textcolor{red}{2} & 0 \\ 2 & 4 & 0 & 7 \\ - 5 & - 1 & 0 & 3\end{matrix}\right)$

we see that the third column has 3 zeroes and only one number: we can cancel the column and line crossing at $2$ (red) to reduce the order of our matrix BUT we have to keep the $2$ we used to do this with the appropriate sign!

Remember that:
$\left(\begin{matrix}+ & - & + & - \\ - & + & \textcolor{red}{-} & + \\ + & - & + & - \\ - & + & - & +\end{matrix}\right)$

so our $2$ must have the corresponding minus sign and we get:
$\textcolor{red}{- 2} \left(\begin{matrix}3 & 5 & 6 \\ 2 & 4 & 7 \\ - 5 & - 1 & 3\end{matrix}\right)$ as reduced matrix!

The determinant can be now "easily" evaluated using Laplace or Sarrus to get:
$\text{Det} = - 2 \left[3 \left(12 + 7\right) - 5 \left(6 + 35\right) + 6 \left(- 2 + 20\right)\right] = - 2 \left[57 - 205 + 108\right] = 80$