# Finding the determinant of a skew-symmetric matrix $K$

Find the determinant of the skew-symmetric matrix $K$

$$K = \begin{bmatrix} 0 & 1 & 3\\ -1 & 0 & 4 \\ -3 & -4 & 0 \\ \end{bmatrix}$$

My Attempted Solution:

I performed the following row operations to reduce $K$ into upper-triangular form $U$

$R_2 \leftrightarrow R_1$

$R_3 - (l_{31} = 3)R_1$

$R_3 \leftrightarrow R_2$

$R_3 - (l_{32} = -4) R_2$

$$U = \begin{bmatrix} -1 & 0 & 4 \\ 0 & -4 & -12 \\ 0 & 0 & -45 \end{bmatrix}$$

From this I got \begin{align} \det(K) &= \pm \ \det(U) \\ &= + \det(U) & \text{(Even no. of row exchanges)} \\ & = (-1)(-4)(-45) \\ &= 180 \end{align}

However the correct answer is $\det(K) = 0$. What could I have done wrong, I wouldn't think it would've been the row operations as the row operations apart from the row exchanges don't affect the $\det(K)$? Any hints or suggestiong are greatly appreciated

## 4 Answers

Your last row operation should have $l_{32}=-\frac14$ instead of $l_{32}=4$.

Also, the determinant of any skew-symmetric matrix with odd dimension must be zero.

Another way to show this is to use the fact that the determinant of a matrix is the product of its eigenvalues, including multiplicities from the characteristic equation. It is also possible to show that all of the eigenvalues of an anti-Hermitian matrix (a generalization of skew-symmetric for complex valued matrices) are purely imaginary. Because all of the elements of this matrix are real, the eigenvalues have to either be 0 or come in complex conjugate pairs. Thus, skew-symmetric matrices can only have a non-zero determinant of the number of dimensions is even.

You can develop the first row to calculate the determinant :

$$\begin{vmatrix} 0 & 1 & 3 \\ -1 & 0 & 4 \\ -3 & -4 & 0 \end{vmatrix}=0-\begin{vmatrix} -1 & 4 \\ -3 & 0 \end{vmatrix}+3\begin{vmatrix} -1 & 0 \\ -3 & -4 \end{vmatrix}=12-12=0.$$

$$\begin{array}{rl} \det \left[\begin{array}{cc|c} 0 & 1 & 3\\ -1 & 0 & 4 \\ \hline -3 & -4 & 0\end{array}\right] &= \det \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} - \det \left(\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} - \begin{bmatrix} 3\\ 4\end{bmatrix} \begin{bmatrix} 3\\ 4\end{bmatrix}^T\right)\\ &= 1 - \det \begin{bmatrix} -9 & -11\\ -13 & -16\end{bmatrix}\\ &= 1 - \det \begin{bmatrix} 9 & 11\\ 13 & 16\end{bmatrix}\\ &= 1 - (\underbrace{144 - 143}_{=1}) = 0\end{array}$$