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

Feb 15, 2016

The determinate of A, |A| = $| A | = 8$

Explanation:

In general you can use "determinant expansion by minors" also or "Laplacian expansion by minors".
- Recipe
$\det \left(A\right) = | A | = {\sum}_{k = 1}^{N} {a}_{i j} {C}_{i j}$
with no implied summation over j and where ${C}_{i j}$ is the cofactor of a_(ij) defined by
C_(ijM_(ij); M_(ij) is Minor of A formed by eliminating the ith row, jth column of A. So expanding around ${a}_{1 j}$, that is eliminating ${a}_{1 j} \mathmr{and} {a}_{i 1}$ we write:

$| A | = {a}_{11} {\left(- 1\right)}^{1 + 1} \textcolor{red}{\det \left(\begin{matrix}{a}_{22} & \cdots & {a}_{2 j} \\ \vdots & \vdots & \vdots \\ {a}_{i 2} & \cdots & {a}_{i j}\end{matrix}\right)} + {\left(- 1\right)}^{1 + 2} {a}_{12} \textcolor{red}{\det \left(\begin{matrix}{a}_{21} & \cdots & {a}_{2 j} \\ \vdots & \vdots & \vdots \\ {a}_{i 2} & \cdots & {a}_{12}\end{matrix}\right)} + \cdots + {\left(- 1\right)}^{1 + j} {a}_{1} j \textcolor{red}{\det \left(\begin{matrix}{a}_{21} & \cdots & {a}_{2 j - 1} \\ \vdots & \vdots & \vdots \\ {a}_{i 2} & \cdots & {a}_{i j - 1}\end{matrix}\right)}$

The reality is you can expand around any row or column so let's be strategic and expand around the 3rd column, but first your signs:
for a 4xx4=>((+,-,+,-), (-,+,-,+),(+,-,+,-), (-,+,-,+)); 3xx3 =>((+,-,+), (-,+,-),(+,-,+));  Note 3rd column has only zero and ones, simplifying life.
$| A | = \cancel{0 \cdot \det \left(\begin{matrix}- 2 & 0 & 0 \\ 0 & - 2 & 1 \\ 0 & - 2 & 0\end{matrix}\right)} - 1 \cdot \det \left(\begin{matrix}- 2 & 0 & - 1 \\ 0 & - 2 & 1 \\ 0 & - 2 & 0\end{matrix}\right) + \cancel{0 \cdot \det \left(\begin{matrix}- 2 & 0 & - 1 \\ - 2 & 0 & 0 \\ 0 & - 2 & 0\end{matrix}\right)} - 1 \cdot \det \left(\begin{matrix}- 2 & 0 & - 1 \\ - 2 & 0 & 0 \\ 0 & - 2 & 1\end{matrix}\right)$
$| A | = 1 \cdot \det \left(\begin{matrix}2 & 0 & 1 \\ 0 & 2 & - 1 \\ 0 & 2 & 0\end{matrix}\right) + 1 \cdot \det \left(\begin{matrix}2 & 0 & 1 \\ 2 & 0 & 0 \\ 0 & 2 & - 1\end{matrix}\right)$
Now if you expand the first term around the first columns and the second term around the 2nd column you will end up:

$| A | = 2 \det \left(\begin{matrix}2 & - 1 \\ 2 & 0\end{matrix}\right) - 2 \det \left(\begin{matrix}2 & 1 \\ 2 & 0\end{matrix}\right)$ cross multiply and subtract
|A| = 2((2*0)- (2*(-1)) -2((2*0)-(2*1)) = 8