Matrix determinants in general

 The determinant of a general $n\times n$ matrix is quite complicated. We give a mechanical, recursive definition first and then think about its meaning later.

Definition (Determinant)

Let $A=(a_{ij})\in M_n$. The determinant $|A|$ of $A$ is defined recursively in the following manner.

• If $n=1$, then $|A|=a_{11}$.
• If $n>1$, then let $A_{ij}$ denote the $(n-1)\times (n-1)$ matrix obtained by removing the $i$-th row and the $j$-th column from $A$, and $$|A|=\sum _{j=1}^n(-1{)}^{i+j}{a}_{ij}|A_{ij}|$$ where $i$ is any arbitrary index from 1 to $n$. (Instead of a row, you may use an arbitrary column to obtain the same result.)

According to this definition, if we want to compute $|A|$ of an $n\times n$ matrix, we need to compute the determinants $|A_{ij}|$ of (many) $(n-1)\times (n-1)$ matrices, which requires computing determinants of $(n-2)\times (n-2)$ matrices, and so on, until we reach the determinants of $1\times 1$ matrices which are trivial to compute (actually we already know how to compute the determinant of a $2\times 2$ matrix so we may stop at $n=2$).

Example. First, let's see the previous definition of the determinant of a $2\times 2$ matrix is consistent with the above definition.

Let $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Then,

$$|A|=(-1{)}^{1+1}a|(d)|+(-1{)}^{1+2}b|(c)|=ad-bc.$$ Thus, the two definitions are indeed consistent. □

Example. 

Let

$$A=\left(\begin{array}{ccc}1& 0& 2\\ -1& 1& 1\\ 0& 1& 0\end{array}\right)=(a_{ij}).$$

To compute its determinants, let's pick the first row as ``$i$''. So

$$\begin{array}{rcl}|A|& =& (-1{)}^{1+1}{a}_{1,1}\left|\begin{array}{cc}{a}_{2,2}& a_{2,3}\\ a_{3,2}& a_{3,3}\end{array}\right|+(-1{)}^{1+2}{a}_{1,2}\left|\begin{array}{cc}{a}_{2,1}& a_{2,3}\\ a_{3,1}& a_{3,3}\end{array}\right|+(-1{)}^{1+3}{a}_{1,3}\left|\begin{array}{cc}{a}_{2,1}& a_{2,2}\\ a_{3,1}& a_{3,2}\end{array}\right|\\ & =& (-1{)}^{1+1}\cdot 1\cdot \left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|+(-1{)}^{1+2}\cdot 0\cdot \left|\begin{array}{cc}-1& 1\\ 0& 0\end{array}\right|+(-1{)}^{1+3}\cdot 2\cdot \left|\begin{array}{cc}-1& 1\\ 0& 1\end{array}\right|\\ & =& -1+0+(-2)=-3.\end{array}$$

Since we may pick any arbitrary row ($i$), we may equally use the second row to compute

$$\begin{array}{rcl}|A|& =& (-1{)}^{2+1}(-1)\left|\begin{array}{cc}0& 2\\ 1& 0\end{array}\right|+(-1{)}^{2+2}(1)\left|\begin{array}{cc}1& 2\\ 0& 0\end{array}\right|+(-1{)}^{2+3}(1)\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|\\ & =& -2+0-1=-3\end{array}$$

to obtain the same result. □

Example. 

Let $X=(x_{ij})\in M_3$. Prove the following.

$$|X|=x_{11}{x}_{22}{x}_{33}-x_{11}{x}_{23}{x}_{32}+x_{12}{x}_{23}{x}_{31}-x_{12}{x}_{21}{x}_{33}+x_{13}{x}_{21}{x}_{32}-x_{13}{x}_{22}{x}_{31}.$$

Do you see any pattern? There are $3!=6$ terms. Each term is a product of three elements. Each of 1, 2, 3 appears only once in the first index of each factor; the same for the second index. Half (3) of the terms are multiplied by -1. What are these patterns? They are permutations! As fascinating as it is, we don't delve into the details here. See a textbook on linear algebra. □