$$

Wrong way to find the determinant

Problem

Let

$$A=\left[\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right].$$

Is

$det(A)$ equal to

$1$ or

$-1$?

Thought

Consider a general matrix

$$M=\left[\begin{array}{cccc}a& b& c& d\\ e& f& g& h\\ i& j& k& \ell \\ m& n& o& p\end{array}\right].$$

It seems nature to calcaulate the determinant by the formulat

$$afkp+bg\ell m+chin+dejo-dgjm-cfip-be\ell o-ahkn.$$

In this sense, the determinant should be

$-1$. However, this is a wrong answer.

The way we calculate the determinant of a

$2\times 2$ or a

$3\times 3$ matrix, such as

$$det\left(\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\right)=ad-bc$$

or

$$det\left(\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\right)=aei+bfg+cdh-ceg-bdi-afh,$$

is called the permutation expansion . If we observe this formula carefully, the determinant of a

$2\times 2$ matrix is the sum of

$2=2!$ terms, and the determinant of a

$3\times 3$ matrix is the sum of

$6=3!$ terms. Indeed, the correct formula for the determinant of a

$4\times 4$ matrix is the sum of

$24=4!$ terms, and the determinant of an

$n\times n$ matrix is the sum of

$n!$ terms. That is why high-school mathematics rarely teach how to find the determinant of a large matrix.

Fortunately, the Laplace expansion learned in high school still works for larger matrix. Let

$T=\left[\begin{array}{c}{t}_{i,j}\end{array}\right]$ be an

$n\times n$ matrix. We may define

$T_{i,j}$ as the

$(n-1)\times (n-1)$ matrix obtained from

$T$ by removing the

$i$-th row and the

$j$-th column. Then, we may expand the determinant along the first row as

$$det(T)=(-1{)}^{1+1}{t}_{1,1}det(T_{1,1})+\cdots +(-1{)}^{1+n}{t}_{1,n}det(T_{1,n}).$$

In general, we may expand the determinant along the

$i$-th row as

$$det(T)=(-1{)}^{i+1}{t}_{i,1}det(T_{i,1})+\cdots +(-1{)}^{i+n}{t}_{i,n}det(T_{i,n}).$$

The expansion along the

$j$-th column is defined similarly.

Sample answer

By Laplace expansion,

$$det(A)=0\cdot \left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right]-0\cdot \left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 1& 0& 0\end{array}\right]+0\cdot \left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]-1\cdot \left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]=1.$$

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