Prove that determinant of skew-symmetric matrix of odd order is zero

Question

Prove that the determinant of a skew-symmetric matrix of odd order is always zero.

(JEE Main 2023)

Solution — Step by Step

A square matrix $A$ is skew-symmetric if $A^T = -A$ . This means every element satisfies $a_{ij} = -a_{ji}$ , and all diagonal elements are zero (since $a_{ii} = -a_{ii} \implies a_{ii} = 0$ ).

For any square matrix: $\det(A^T) = \det(A)$ .

Since $A^T = -A$ for a skew-symmetric matrix:

\det(A^T) = \det(-A)

\det(A) = \det(-A)

For an $n \times n$ matrix: $\det(kA) = k^n \det(A)$ .

So $\det(-A) = (-1)^n \det(A)$ .

Substituting back:

\det(A) = (-1)^n \det(A)

If $n$ is odd, then $(-1)^n = -1$ :

\det(A) = -\det(A)

2\det(A) = 0

\boxed{\det(A) = 0}

For even $n$ , we get $\det(A) = \det(A)$ , which is always true — so the determinant need not be zero for even-order skew-symmetric matrices.

Why This Works

The proof uses two fundamental properties of determinants: (1) transposing doesn’t change the determinant, and (2) multiplying a matrix by a scalar $k$ multiplies the determinant by $k^n$ . Combining these with the skew-symmetric condition $A^T = -A$ forces the determinant to equal its own negative — which is only possible if the determinant is zero.

Notice the proof fails for even order because $(-1)^n = 1$ when $n$ is even, giving the trivially true $\det(A) = \det(A)$ . Indeed, the $2 \times 2$ skew-symmetric matrix $\begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}$ has determinant $a^2$ , which is generally non-zero.

Alternative Method — Direct verification for 3x3

For a $3 \times 3$ skew-symmetric matrix:

A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}

Expanding along the first row: $\det(A) = 0 \cdot M_{11} - a(0 - (-bc)) + b((-a)(-c) - 0)$

$= -a(bc) + b(ac) = -abc + abc = 0$ .

In JEE Main, this is a favourite 1-mark MCQ: “If $A$ is a $3 \times 3$ skew-symmetric matrix, then $\det(A) = ?$ ” Answer: 0. Know the proof — but also know that for even order, the answer is NOT necessarily zero. JEE loves testing whether you remember the “odd order” condition.

Common Mistake

Students sometimes claim “determinant of ALL skew-symmetric matrices is zero.” This is false — it’s only true for odd order. A $2 \times 2$ or $4 \times 4$ skew-symmetric matrix can have a non-zero determinant. The odd-order condition is essential, and forgetting it in an MCQ leads to marking the wrong option.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct verification for 3x3

Common Mistake

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