ABᵀ – BᵀA is Skew-Symmetric

📘 Question

If \(A\) and \(B\) are matrices of suitable order, then:

\[ AB^T – B^T A \]

is a:

(a) skew-symmetric matrix
(b) null matrix
(c) unit matrix
(d) symmetric matrix


✏️ Step-by-Step Solution

Step 1: Take transpose

\[ (AB^T – B^T A)^T = (AB^T)^T – (B^T A)^T \]

Step 2: Use transpose property

\[ (AB^T)^T = B A^T,\quad (B^T A)^T = A^T B \]
\[ (AB^T – B^T A)^T = BA^T – A^T B \]

Step 3: Rearrange

\[ = -(AB^T – B^T A) \]

Step 4: Conclusion

Since:

\[ M^T = -M \]

Matrix is skew-symmetric.


✅ Final Answer

\[ \boxed{(a)\; \text{skew-symmetric matrix}} \]

💡 Key Concept

A matrix is skew-symmetric if:

\[ A^T = -A \]

Use transpose of product:

\[ (AB)^T = B^T A^T \]

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