Determinant of order 4 question I don't get this, maybe you can explain me how you solve it
"Write down all the terms appearing in the determinant of order 4 which have a minus sign and contain the factor a23"
Thanks a lot
 A: One definition of the determinant of an $n\times n$ matrix is:
$$\det(A) = \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)},$$
where $\sigma$ runs over all permutations of $S_n$.
You are being asked to write down all those in which $\mathrm{sgn}(\sigma)=-1$, and $\sigma(2)=3$.
For example, the same question with a $3\times 3$ determinant: the permutations that have $\sigma(2)=3$ are $\sigma=(2,3)$ and $\sigma=(1,2,3)$. The first one is odd, the second one is even, so the only term with negative sign would be
$$\mathrm{sgn}(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)} = -a_{11}a_{23}a_{32}.$$
If your definition of determinant is by using minors/cofactors, then consider expanding along the second row (or the third column), so that the only terms that contain the factor $a_{23}$ are those in which you compute the minor of $a_{23}$. Keep track of the sign and do the $3\times 3$ determinant of the minor to figure out which terms satisfy the desired conditions.
Again, using a $3\times 3$ example,
$$A = \left(\begin{array}{ccc}
a_{11} &a_{12} &a_{13}\\
a_{21} & a_{22} &a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array}\right),$$
expanding along the third column, the only terms with a factor of $a_{23}$ are the ones that come from
$$-a_{23}\left|\begin{array}{cc}
a_{11}&a_{12}\\
a_{31}&a_{32}
\end{array}\right|  = -a_{23}\Bigl( a_{11}a_{32} - a_{12}a_{31}\Bigr).$$
So the only one with a negative sign is $-a_{11}a_{23}a_{32}$, same answer as above.
