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# Wrong way to find the determinant

## Problem

Let

$$
A = \begin{bmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix}.
$$

Is $\det(A)$ equal to $1$ or $-1$?

## Thought

Consider a general matrix

$$
M = \begin{bmatrix}
a & b & c & d \\
e & f & g & h \\
i & j & k & \ell \\
m & n & o & p
\end{bmatrix}.
$$

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

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

or

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

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 $T = \begin{bmatrix} t_{i,j} \end{bmatrix}$ 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,

$$
\det(A) =
0\cdot\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix} -
0\cdot\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix} +
0\cdot\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{bmatrix} -
1\cdot\begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{bmatrix} =
1.
$$

*This note can be found at Course website > Learning resources.*