Q.$\begin{vmatrix} 0 & x - y & x - z \\ y - x & 0 & y - z \\ z - x & z - y & 0 \end{vmatrix} =$ ________ .

Concept: Determinant Evaluation Using Identities — notice the skew-symmetric pattern: each off-diagonal entry is the negative of its mirror across the diagonal.

Step 1: Observe that the matrix is skew-symmetric: $a_{ij} = -a_{ji}$ for all $i,j$, and all diagonal entries are $0$. …

Why This Approach Works

When you see a matrix where the diagonal is all zeros and the off-diagonal entries are negatives of each other (like $x-y$ and $y-x$), you're looking at a skew-symmetric matrix. For any skew-symmetric matrix of odd order, the determinant is identically zero — no matter what the variables are. This isn't a coincidence; it's a fundamental property that saves you from messy algebra.

Let's verify this step by step, and also see what happens if you try to expand directly.

Step-by-Step Solution

1. Identify the matrix type

The given matrix is:

Notice that each entry $a_{ij}$ satisfies $a_{ij} = -a_{ji}$. For example:

• $a_{12} = x - y$ and $a_{21} = y - x = -(x - y)$
• $a_{13} = x - z$ and $a_{31} = z - x = -(x - z)$
• $a_{23} = y - z$ and $a_{32} = z - y = -(y - z)$

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

2. Apply the key property

Here $n = 3$, which is odd. Therefore, $\Delta = 0$ directly.

3. Why does this property hold? (Quick proof)

Take the determinant of both sides of $A^T = -A$:

We know $\det(A^T) = \det(A)$. Also, $\det(-A) = (-1)^n \det(A)$ because multiplying each of the $n$ rows by $-1$ multiplies the determinant by $(-1)^n$.

So:

If $n$ is odd, $(-1)^n = -1$, giving $\det(A) = -\det(A)$, which forces $\det(A) = 0$.

4. Verification by direct expansion (optional)

If you prefer to see it with algebra, expand along the first row: …

Method: Recognising and Using the Skew-Symmetric Determinant Shortcut

This method applies whenever a determinant's diagonal entries are all zero and every off-diagonal pair of entries are negatives of each other — instead of expanding, use the skew-symmetry property directly.

Steps

Step 1: Check the pattern

Look at the matrix entries. If $a_{ii}=0$ for every $i$, and $a_{ij} = -a_{ji}$ for every pair $i \ne j$, the matrix is skew-symmetric ($A^T = -A$).

Step 2: Check the order (size) of the matrix

Count the order $n$ of the square matrix. The shortcut only gives an instant zero when $n$ is odd.

Step 3: Apply the odd-order skew-symmetric rule

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

Common Mistakes

Mistake 1: Expanding the determinant term by term instead of spotting the pattern

Why it's wrong: a full cofactor expansion here is long and error-prone (multiple sign flips across binomial-looking entries), when the skew-symmetric shortcut gives the answer instantly. Correct approach: first check whether the diagonal is all zeros and off-diagonal entries are mirror-negatives — if so, use the odd-order shortcut instead of expanding.

Mistake 2: Applying the "always zero" shortcut regardless of order …