# 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

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Consider the determinant $$\begin{vmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&‌​a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}\end{vmatrix}$$ ; if you expand it out, you should see terms that fit the criteria in your question... – J. M. Oct 14 '11 at 17:19
What is your definition of "determinant"? – Arturo Magidin Oct 14 '11 at 17:19

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.

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