Evaluate
\[\begin{vmatrix} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{vmatrix}.\]
We can expand the determinant as follows:
\begin{align*}
\begin{vmatrix} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{vmatrix} &= -\sin \alpha \begin{vmatrix} -\sin \alpha & \sin \beta \\ \cos \alpha & 0 \end{vmatrix} - \cos \alpha \begin{vmatrix} -\sin \alpha & 0 \\ \cos \alpha & -\sin \beta \end{vmatrix} \\
&= -\sin \alpha (-\sin \beta \cos \alpha) - \cos \alpha (\sin \alpha \sin \beta) \\
&= \boxed{0}.
\end{align*}
0
Level 1
Precalculus
Problem: Evaluate
\[\begin{vmatrix} 0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0 \end{vmatrix}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate the determinant of the given 3x3 matrix, we can use the formula for the determinant of a 3x3 matrix, which is given by:

\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]

where the matrix is represented as:

\[
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\]

In our case, we identify the elements of the matrix as follows:

- \( a = 0 \), \( b = \sin \alpha \), \( c = -\cos \alpha \)
- \( d = -\sin \alpha \), \( e = 0 \), \( f = \sin \beta \)
- \( g = \cos \alpha \), \( h = -\sin \beta \), \( i = 0 \)

Substituting these values into the determinant formula, we have:

\[
\text{det}(A) = 0(0 \cdot 0 - \sin \beta \cdot (-\sin \beta)) - \sin \alpha(-\sin \alpha \cdot 0 - \sin \beta \cdot \cos \alpha) + (-\cos \alpha)(-\sin \alpha \cdot (-\sin \beta) - 0 \cdot \cos \alpha)
\]

This simplifies to:

\[
\text{det}(A) = 0 - \sin \alpha(-\sin \beta \cos \alpha) + (-\cos \alpha)(\sin \alpha \sin \beta)
\]

Both terms involving \(\sin \alpha\) and \(\cos \alpha\) will cancel each other out:

\[
\text{det}(A) = \sin \alpha \sin \beta \cos \alpha - \sin \alpha \sin \beta \cos \alpha = 0
\]

Thus, the determinant evaluates to zero. The reasoning shows that the contributions from the terms involving \(\sin \alpha\) and \(\cos \alpha\) negate each other, leading to the conclusion that the determinant of the matrix is zero.<|END_OF_LATENT|> The final answer is $0$.