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If P is a skew-symmetric matrix of order 3, then what is det(P) equal to?

This question was previously asked in
NDA 2 2024 GAT Question Paper (01-Sep-2024)
The correct answer is
0

Skew-Symmetric Matrix Determinant Explained

A square matrix 'P' is defined as skew-symmetric if its transpose equals its negative. In mathematical terms, this property is expressed as:

\( P^T = -P \)

This definition implies that for any entry \(p_{ij}\) at row \(i\) and column \(j\), the entry \(p_{ji}\) at row \(j\) and column \(i\) must satisfy \(p_{ji} = -p_{ij}\). A direct consequence is that all elements on the main diagonal of a skew-symmetric matrix must be zero. This is because for a diagonal element \(p_{ii}\), the condition becomes \(p_{ii} = -p_{ii}\), which simplifies to \(2p_{ii} = 0\), meaning \(p_{ii} = 0\).

Properties of Skew-Symmetric Matrix Determinants

To determine the value of the determinant of a skew-symmetric matrix, we utilize fundamental properties of determinants:

  • The determinant of a matrix is invariant under transposition: \(\det(P^T) = \det(P)\).
  • When a matrix is multiplied by a scalar \(k\), its determinant is multiplied by \(k^n\), where \(n\) is the order (size) of the matrix: \(\det(kA) = k^n \det(A)\).

Using the definition \(P^T = -P\), we can equate the determinants:

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

Applying the properties mentioned above, we substitute \(\det(P^T)\) with \(\det(P)\) and \(\det(-P)\) with \((-1)^n \det(P)\):

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

Applying the Property to Order 3

The specific question concerns a skew-symmetric matrix P of order 3. Thus, the size of the matrix is \(n = 3\). Since 3 is an odd number, the term \((-1)^n\) becomes \((-1)^3\), which equals -1.

Substituting \(n=3\) into the equation \(\det(P) = (-1)^n \det(P)\), we get:

\( \det(P) = (-1)^3 \det(P) \)

\( \det(P) = -1 \cdot \det(P) \)

\( \det(P) = -\det(P) \)

To find the value of \(\det(P)\), we can rearrange the equation:

\( \det(P) + \det(P) = 0 \)

\( 2 \det(P) = 0 \)

Finally, dividing by 2 yields:

\( \det(P) = \frac{0}{2} \)

\( \det(P) = 0 \)

This result demonstrates a key principle: the determinant of any skew-symmetric matrix with an odd order is always zero.

Matrix Order (n) Determinant Equation Result for det(P)
Odd (e.g., 3) \(\det(P) = (-1)^n \det(P)\) implies \(\det(P) = -\det(P)\) \(\det(P) = 0\)
Even (e.g., 2, 4) \(\det(P) = (-1)^n \det(P)\) implies \(\det(P) = \det(P)\) Can be non-zero

Therefore, for the given matrix P, which is skew-symmetric and of order 3, its determinant is 0.

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