Question: If $A$ and $B$ are skew-symmetric matrices of same order, then $AB' + BA'$ is a/an :

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

Given that $A$ and $B$ are skew-symmetric matrices, by definition:

• $A' = -A$
• $B' = -B$

A matrix $X$ is symmetric if $X' = X$ and skew-symmetric if $X' = -X$. We need to check the transpose of the expression $X = AB' + BA'$.

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

Let $X = AB' + BA'$. Taking the transpose of both sides:

$$X' = (AB' + BA')'$$

Using the properties of transpose: $(P + Q)' = P' + Q'$ and $(PQ)' = Q'P'$:

$$X' = (AB')' + (BA')'$$ $$X' = (B')'A' + (A')'B'$$

Since $(M')' = M$, we substitute the values:

$$X' = BA' + AB'$$

Substitute the skew-symmetric properties ($A' = -A$ and $B' = -B$):

$$X' = B(-A) + A(-B)$$ $$X' = -BA - AB$$ $$X' = -(AB + BA)$$

Step 3 (Final Concluded Answer): [1/3 Mark]

Since $X' = -(AB + BA)$, and given $A' = -A$ and $B' = -B$, the expression $AB' + BA'$ becomes $A(-B) + B(-A) = -(AB + BA)$.

Thus, $X' = -X$. Therefore, the matrix is a skew-symmetric matrix.

Correct Option: (B)

💡 EL10 Concept Insight: Think of a skew-symmetric matrix as a "negative mirror" of itself; when you combine two of these mirrors in this specific pattern, the negative signs flip the whole result back to its negative, which is exactly what makes a matrix skew-symmetric!