Question

If $A,B$ be the two symmetric matrices of order $3$.
Statement-$1$
$A(BA)$ and $(AB)A$ are symmetric matrices.
Statement-$2$
$AB$ is a symmetric matrix if $A,B$ multiplication is commutative.
A. Statement-$1$ is true, Statement-$2$ is true; Statement-$2$ is correct explanation for Statement-$1$
B. Statement-$1$ is true, Statement-$2$ is true; Statement-$2$ is not a correct explanation for Statement-$1$
C. Statement-$1$ is true, Statement-$2$ is false.
D. Statement-$1$ is false, Statement-$2$ is true

Answer 1

In this question, we are given that $A,B$ are the two symmetric matrices of order$3$. Now we will use the definition of symmetric matrices to check whether statement (1) and (2) are true or not.

Complete step-by-step answer:
In this question, we are given that $A,B$ are the two symmetric matrices of order $3$.
And we know that $M$ is known as the symmetric matrix if ${M^T} = M$, which means that the transpose of $M$ is also $M$.
So from this definition, we can say that
${A^T} = A$ $- - - - - (1)$
${B^T} = B$ $- - - - - (2)$
Now according to the question we are given two statements, (1) and (2).
Firstly we will see whether the statement (1) is true or not.
Statement (1) says that
$A(BA)$ and $(AB)A$ are symmetric matrices.
Let us first consider ${[A(BA)]^T}$
We know that if $M,N$ are two matrices then ${(MN)^T} = {N^T}{M^T}$
So using this in the statement (1), we get
${[A(BA)]^T} = {(BA)^T}{A^T}$
We know that if $M,N$ are two matrices then ${(MN)^T} = {N^T}{M^T}$
So
${[A(BA)]^T} = {A^T}{B^T}{A^T}$
Now substituting the value from (1) and (2) of ${A^T}$and ${B^T}$
${[A(BA)]^T} = A.B.A$ $- - - - (3)$
We get that$A(BA)$ is a symmetric matrix.
Now we will consider $[(AB)A){]^T}$
$[(AB)A){]^T}$ $= {A^T}{(AB)^T}$
$[(AB)A){]^T}$ $= {A^T}{B^T}{A^T}$
Now again substituting the values from (1) and (2) of ${A^T}$and ${B^T}$
$[(AB)A){]^T}$ $= A.B.A$ $- - - - - (4)$
So from (3) and (4), we can conclude that statement 1 is the correct statement.

Now we will check for the statement (2)
Statement (2) says that $AB$ is a symmetric matrix if $A,B$ multiplication is commutative.
If multiplication of $AB$ is commutative and it is a symmetric matrix, then we know that
$AB = BA$ $- - - - (5)$
Now taking transpose on both the sides
${(AB)^T} = {(BA)^T}$
${(AB)^T}$ $= {A^T}{B^T}$ $- - - - (6)$
Now substituting the value from (1) and (2) of ${A^T}$ and ${B^T}$ in (6)
${(AB)^T}$ $= AB$
As its transpose is to itself, so we can say that $AB$ is a symmetric matrix.
Hence statement (2) is also correct.
If we clearly observe, we did not use statement (2) to prove the statement (1) and vice-versa.
Hence they are not the correct explanation of each other.

So, the correct answer is “Option B”.

Note: We can solve statement (2) by the alternate method also.
Statement (2) says that $AB$ is a symmetric matrix if $A,B$ multiplication is commutative.
So we can write ${(AB)^T}$ $= {B^T}{A^T}$
Now again substituting the values from (1) and (2)
We get that ${(AB)^T}$ $= BA$
Now if the multiplication is given commutative, then $AB = BA$
We get ${(AB)^T}$ $= BA$ $= AB$
Which implies that $AB$ is also symmetric.