How to Find the Determinant of a Skew-Symmetric Matrix

A matrix is skew-symmetric if transposing it changes every entry’s sign: $A^{T}=-A$. The question does not specify the matrix entries, so there is no single numerical determinant to calculate. But the order of a skew-symmetric matrix gives an immediate result when it is odd, and a useful Pfaffian formula handles even orders.

These facts apply to matrices over the real or complex numbers. More generally, the odd-order argument works over a field whose characteristic is not 2.

The determinant rule for odd order

Suppose $A$ is an $n\times n$ skew-symmetric matrix. A matrix and its transpose have the same determinant, while multiplying an $n\times n$ matrix by $-1$ multiplies its determinant by $(-1)^n$. Therefore,

$$\begin{aligned} \det(A) &= \det(A^T) \\ &= \det(-A) \\ &= (-1)^n\det(A). \end{aligned}$$

When $n$ is odd, $(-1)^n=-1$, so this says $\det(A)=-\det(A)$. Over the real or complex numbers, that forces $\det(A)=0$.

For example, any $3\times3$ skew-symmetric matrix has the form

$$A=\begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix},$$

and its determinant is zero, regardless of the values of $a$, $b$, and $c$. The same result holds for every odd order, not just 3.

What happens for even order?

For even order, the sign argument gives no conclusion: $(-1)^n=1$, so it only says $\det(A)=\det(A)$. The determinant can be zero or nonzero. For instance,

$$\begin{pmatrix}0&a\\-a&0\end{pmatrix}$$

has determinant $a^2$, which is nonzero when $a\ne0$.

For even-order skew-symmetric matrices, the general identity is

$$\det(A)=\operatorname{Pf}(A)^2,$$

where $\operatorname{Pf}(A)$ is the Pfaffian. It is a polynomial in the entries above the diagonal. For a $4\times4$ matrix

$$A=\begin{pmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0 & f \\ -c & -e & -f & 0 \end{pmatrix},$$

the Pfaffian is $af-be+cd$, so

$$\det(A)=(af-be+cd)^2.$$

This formula is often quicker than expanding the determinant by minors.

Worked 4 × 4 example

Consider

$$A=\begin{pmatrix} 0 & 2 & -1 & 3 \\ -2 & 0 & 4 & 1 \\ 1 & -4 & 0 & 5 \\ -3 & -1 & -5 & 0 \end{pmatrix}.$$

It is skew-symmetric: its diagonal entries are zero, and each entry below the diagonal is the negative of the corresponding entry above it. Match its entries to the 4 × 4 pattern: $a=2$, $b=-1$, $c=3$, $d=4$, $e=1$, and $f=5$. Then

$$\begin{aligned} \operatorname{Pf}(A) &= af-be+cd \\ &= (2)(5)-(-1)(1)+(3)(4) \\ &= 23. \end{aligned}$$

Thus

$$\det(A)=23^2=529.$$

The determinant is positive here because it is the square of a real number.

Common mistakes

• Assuming every skew-symmetric matrix has determinant zero. That is guaranteed only for odd order. The 2 × 2 example above has determinant $a^2$.
• Using the odd-order proof for an even-order matrix. The proof relies on $(-1)^n=-1$, which is true only when $n$ is odd.
• Forgetting the signs in the 4 × 4 Pfaffian. The order is $af-be+cd$, not a sum of all three products. Substitute the signed matrix entries as written.
• Treating “skew-symmetric” as enough information for a numerical answer. It determines the odd-order result, but an even-order determinant depends on the actual entries.

If a particular structured matrix was intended, its entries or defining pattern are needed to compute its specific determinant. The odd/even rule and, for a 4 × 4 matrix, the Pfaffian formula still provide a useful first check.