# Determinant of a matrix

Having some problems with a determinant of a 4x4 matrix M.

$M = \left( {\begin{array}{cc} 1 & 2 & 3 &-1 \\ 0 & 1 & 2 & 2 \\ 1 &1 &0 &0 \\ 3&1&2&0 \end{array} } \right)$

Went along and developed it according to the 4th column. So I end up with two matrixes A and B.

$A = -1 \cdot det \left( {\begin{array}{cc} 0 & 1 & 2 \\ 1 & 1 & 0 \\ 3 &1 &2 \\ \end{array} } \right)$

$B = 2 \cdot det \left( {\begin{array}{cc} 1 & 2 & 3 \\ 1 & 1 & 0 \\ 3 &1 &2 \\ \end{array} } \right)$

I get $A= (-1) \cdot((0 \cdot1\cdot2)+(1\cdot0\cdot3)+(2\cdot1\cdot1)-(3\cdot1\cdot2)-(1\cdot2\cdot2)-(1\cdot1\cdot0)) \\$

$A=(-1) \cdot(-6)=6$

$B= 2 \cdot((1\cdot1\cdot2)+(2\cdot0\cdot3)+(3\cdot1\cdot1)-(3\cdot1\cdot3)-(1\cdot2\cdot2)-(1\cdot1\cdot0)) \\$

$B = 2\cdot8=16$

$A+B=22$

which is wrong. Where is my mistake? The correct answer should be $-22$ but I don't get why my solution keeps being positive.

Edit: im such a moron: A = 1* det and B = -2 * det. Everythings clearing up while in bed. Hehe!

• You need to alternate the sign with each element, starting with '+' for the top left element. Jun 11, 2013 at 18:37

The formula to calculate the determinant is given by

$$\sum_{\sigma \in S_n} (-1)^{\sigma} a_{1\sigma(1)} \times \cdots \times a_{n \sigma(n)},$$

if you want me to explain the whole formula, I can but the important bit right now is the $(-1)^{\sigma}$ which refers to the sign of each entry you are expanding by.

First, you have expanded by element $a_{1,4}$ and so the sign should be $(-1)^{1 + 4} = -1$ and so your determinant of $A$ should be

$$(-1) \cdot (-1) \cdot (-6) = -6.$$

So, your sign for when expanding by row $4$ column $2$ is going to be $(-1)^6 = 1$.

Another mistake you have made is that the determinant of $B$ isn't $16$, it's $-16$.

• Ok, re-reading the question, it's not exactly "determinant of $A$" and "determinant of $B$", but you get what I mean? Jun 11, 2013 at 18:47

You are expanding down the fourth column, which means that the cofactor on entry $\ a_{14} = -1 \$ is negative and the cofactor on entry $\ a_{24} = 2 \$ is positive. You should have

$$[ (-1) \cdot (-1) \cdot ( 2 - 6 - 2 ) ] \ + \ [ (+1) \cdot 2 \cdot ( 2 + 3 - 9 - 4 ) ]$$

$$= \ (-6) \ + \ (-16) \ = \ -22 \ .$$ Your mistake was that you forgot the $(-1)^{i+j}$ for the cofactors.