Q.Evaluate $\Delta = \begin{vmatrix} 0 & \sin\alpha & -\cos\alpha \\ -\sin\alpha & 0 & \sin\beta \\ \cos\alpha & -\sin\beta & 0 \end{vmatrix}$.

Concept: Determinant Evaluation Using Identities — here, the matrix is skew-symmetric of odd order, so its determinant is zero.

Step 1: Check the property. A square matrix $A$ is skew-symmetric if $A^T = -A$. For the given matrix, each entry satisfies $a_{ij} = -a_{ji}$, and the diagonal entries are all zero. …

Why This Approach Works

When you see a determinant with zeros on the main diagonal and opposite signs across the diagonal, you're looking at a skew-symmetric matrix. For any skew-symmetric matrix of odd order, the determinant is always zero. This isn't a coincidence — it follows directly from the property $A^T = -A$ and the fact that $\det(A^T) = \det(A)$.

Let's see why.

Step-by-Step Solution

1. Identify the matrix type

The given matrix is:

Notice:

• Every diagonal entry is $0$.
• For any $i \neq j$, the entry at $(i,j)$ is the negative of the entry at $(j,i)$.

This is exactly the definition of a skew-symmetric matrix: $A^T = -A$.

2. Apply the determinant property

For any square matrix $A$, we know $\det(A^T) = \det(A)$.

If $A$ is skew-symmetric, then $A^T = -A$, so:

3. Factor out the scalar

For a $3 \times 3$ matrix, $\det(-A) = (-1)^3 \det(A) = -\det(A)$.

Thus:

4. Solve the equation

The only number equal to its own negative is zero:

$\det(A) = 0$ …

Method: Recognizing a Skew-Symmetric Determinant of Odd Order

This method evaluates a determinant instantly, without any expansion, by recognizing that it belongs to a special structural class: a skew-symmetric matrix of odd order.

Steps

Step 1: Check the diagonal

A skew-symmetric matrix always has every diagonal entry equal to zero (since $a_{ii}=-a_{ii}$ forces $a_{ii}=0$).

Step 2: Check the off-diagonal pairing

Confirm that for every pair of positions, $a_{ij}=-a_{ji}$ — i.e. the matrix equals the negative of its own transpose, $A^T=-A$.

Step 3: Check the order (size) of the matrix

This shortcut only applies when the matrix is odd-order ($3\times3$, $5\times5$, etc.) — the argument breaks down for even order.

Step 4: Apply the determinant property

Since $\det(A^T)=\det(A)$ always, and $A^T=-A$ here, substitute:

$\det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)$ …

Common Mistakes

Mistake 1: Trying to expand the determinant directly instead of recognizing the skew-symmetric pattern

Why it's wrong: direct expansion works but is slower and riskier — it's easy to make a sign error among the several $\sin\alpha,\cos\alpha,\sin\beta$ terms, when the entire determinant is actually forced to be zero by structure alone. Correct approach: first check the diagonal (all zero) and the off-diagonal pairing ($a_{ij}=-a_{ji}$) — if both hold and the order is odd, the determinant is $0$ with no further computation needed.

Mistake 2: Applying the "skew-symmetric ⇒ determinant zero" shortcut to an even-order matrix …