# How do you find the determinant of ((0, 1, 2, 3), (0, 0, 2, 0), (3, 10, 12, 20), (0, 0, 0, 4))?

Feb 16, 2016

given by

$- 2 \det \left(\begin{matrix}0 & 1 & 3 \\ 3 & 10 & 20 \\ 0 & 0 & 4\end{matrix}\right)$

and

$\det \left(\begin{matrix}0 & 1 & 3 \\ 3 & 10 & 20 \\ 0 & 0 & 4\end{matrix}\right)$

is given by

$4 \det \left(\begin{matrix}0 & 1 \\ 3 & 10\end{matrix}\right)$

and

$\det \left(\begin{matrix}0 & 1 \\ 3 & 10\end{matrix}\right)$

is given by

$0 \cdot 10 - 3 \cdot 1 = - 3$

In fact you can develop on what line or column you want, find one which have most $0$

Then remember formula

$\det \left(A\right) = {\sum}_{n = 0}^{n} {\left(- 1\right)}^{i + \left(j + n\right)} {a}_{i \left(j + n\right)} \det \left({A}_{i \left(j + n\right)}\right)$

Complex ? not really

here you develop the determinant in line because $i$ is fixed (and $= 2$, you just have to read the number which is ${a}_{i j}$ and to have ${A}_{i j}$ just hide with your finger the column and the line which contain ${a}_{i j}$

here it's simple, ${a}_{21} = {a}_{22} = {a}_{24} = 0$ so you just have

$\det \left(A\right) = {\left(- 1\right)}^{2 + 3} {a}_{23} \det \left({A}_{23}\right)$

repeat that until you have $2 \times 2$ matrix like i did