<s>[INST] 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
 [/INST] You can develop the first row to calculate the determinant :
$$\begin{vmatrix} 0 & 1 & 3 \\ -1 & 0 & 4 \\ -3 & -4 & 0 \end{vmatrix}=0-\begin{vmatrix} -1 & 4 \\ -3 & 0 \end{vmatrix}+3\begin{vmatrix} -1 & 0 \\ -3 & -4 \end{vmatrix}=12-12=0.$$
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