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If A and B are matrices of same order, then (ABT−BAT)(AB^T - BA^T) is always

Solution

✅ Correct Option: 2

Let P=ABT−BATP = AB^T - BA^T

To determine the nature of this matrix, we find PTP^T and compare it with PP.


PT=(ABT−BAT)TP^T = (AB^T - BA^T)^T

PT=(ABT)T−(BAT)TP^T = (AB^T)^T - (BA^T)^T

Using (XY)T=YTXT(XY)^T = Y^TX^T:

PT=(BT)TAT−(AT)TBTP^T = (B^T)^TA^T - (A^T)^TB^T


Using (XT)T=X(X^T)^T = X:

PT=BAT−ABTP^T = BA^T - AB^T

PT=−(ABT−BAT)P^T = -(AB^T - BA^T)

PT=−PP^T = -P


Since PT=−PP^T = -P, the matrix (ABT−BAT)(AB^T - BA^T) is always a skew-symmetric matrix.

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