Q.If $A$ and $B$ are symmetric matrices of the same order, then show that $AB$ is symmetric if and only if $A$ and $B$ commute, that is $AB = BA$.

Concept: Symmetric And Skew Symmetric Matrices

A matrix $P$ is symmetric if $P^T = P$. We use the property $(AB)^T = B^T A^T$ and the given symmetry of $A$ and $B$.

Step 1: Since $A$ and $B$ are symmetric, $A^T = A$ and $B^T = B$.

Step 2: Compute $(AB)^T$:

$(AB)^T = B^T A^T = B A$. …

Why this works — the core idea

A symmetric matrix equals its own transpose. So when you multiply two symmetric matrices $A$ and $B$, the transpose of the product is $(AB)^T = B^T A^T$. Because $A$ and $B$ are symmetric, $B^T = B$ and $A^T = A$, giving $(AB)^T = BA$.

Now here's the punchline: for $AB$ to be symmetric, we need $(AB)^T = AB$. That forces $BA = AB$ — exactly the condition that $A$ and $B$ commute. The whole proof is just this one transpose property, applied twice (once for each direction).

Step-by-step proof

1. The forward direction: If $AB$ is symmetric, then $AB = BA$

We are given that $A$ and $B$ are symmetric, so $A^T = A$ and $B^T = B$.

Since $AB$ is symmetric, we have $(AB)^T = AB$.

But $(AB)^T = B^T A^T$ by the reversal property of transposes. Substituting the symmetry of $A$ and $B$:

Therefore $AB = BA$, which is exactly what we needed to show.

2. The backward direction: If $AB = BA$, then $AB$ is symmetric

Again, $A^T = A$ and $B^T = B$.

Take the transpose of $AB$:

But we are given that $AB = BA$. So $(AB)^T = BA = AB$.

Since $(AB)^T = AB$, the matrix $AB$ is symmetric by definition. …

Method: Proving an "if and only if" using transpose properties

Use this reasoning pattern for statements like "$AB$ is symmetric iff $A$ and $B$ commute", where $A,B$ are symmetric.

Steps

Step 1: Translate "symmetric" into a transpose equation and prove one direction.

$M$ symmetric means $M' = M$. Assume $AB$ is symmetric: $(AB)' = AB$. Using the reversal law and $A'=A$, $B'=B$, get $(AB)' = B'A' = BA$, so $AB = BA$.

Step 2: Prove the converse. …

Common Mistakes

Mistake 1: Assuming the product of symmetric matrices is always symmetric.

Why it's wrong: it is symmetric only when the two commute; a non-commuting pair of symmetric matrices gives a non-symmetric product. Correct approach: use $(AB)' = BA$ and require $AB = BA$.

Mistake 2: Using $(AB)' = A'B'$ instead of $B'A'$.

Why it's wrong: the transpose of a product reverses the order; this reversal is exactly what produces $BA$. Correct approach: apply $(AB)' = B'A'$. …