Q.If $A$, $B$ are symmetric matrices of same order, then $AB - BA$ is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix

The key idea is that for symmetric matrices $A = A^T$ and $B = B^T$, the commutator $AB - BA$ is always skew-symmetric.

Step 1: Take the transpose of the given expression:

$(AB - BA)^T = (AB)^T - (BA)^T$.

Step 2: Use the reversal property $(XY)^T = Y^T X^T$ and the symmetry of $A$ and $B$:

$(AB)^T = B^T A^T = BA$, and $(BA)^T = A^T B^T = AB$.

Step 3: Substitute back:

$(AB - BA)^T = BA - AB = -(AB - BA)$. …

Why this works: the idea of symmetry and swapping

A symmetric matrix equals its own transpose: $A^T = A$. A skew-symmetric matrix is the opposite: $S^T = -S$. The key insight is that when you multiply two symmetric matrices, the product $AB$ is not necessarily symmetric — but its transpose is $B^TA^T = BA$. So $AB$ and $BA$ are transposes of each other.

Now look at $AB - BA$. If you take its transpose, you swap the order and get $BA - AB$, which is exactly the negative of what you started with. That's the defining property of a skew-symmetric matrix.

Step-by-step reasoning

1. Write what we know about $A$ and $B$.

Since both are symmetric of the same order:

2. Take the transpose of $AB - BA$.

Remember: $(XY)^T = Y^T X^T$. So:

3. Substitute the symmetric property.

Replace $A^T$ with $A$ and $B^T$ with $B$:

4. Compare with the original expression.

Notice that $BA - AB$ is exactly $-(AB - BA)$. So:

$(AB - BA)^T = -(AB - BA)$ …

Method: Classifying a matrix built from symmetric matrices via its transpose

To decide whether an expression like $AB-BA$ is symmetric or skew symmetric, take its transpose using the reversal law and the given symmetry, then compare with the original.

Steps

Step 1: Transpose the expression

Use $(AB-BA)'=(AB)'-(BA)'=B'A'-A'B'$.

Step 2: Apply the symmetry of $A$ and $B$ …

Common Mistakes

Mistake 1: Assuming $AB$ is symmetric because $A$ and $B$ are

Why it's wrong: $(AB)'=B'A'=BA$, which equals $AB$ only if $A$ and $B$ commute; in general $AB$ is not symmetric. Correct approach: transpose the whole expression $AB-BA$ rather than judging $AB$ alone.

Mistake 2: Choosing "zero matrix" as the answer …