Question

Let $ A $ and $ B $ be two matrices. Prove that $ AB = BA $ if and only if $ AB $ is a symmetric matrix.

Answer 1

Hint: So in this question, we will use the concept of a symmetric matrix. And we know that the symmetric matrix is always the square matrix. And it will always be equal to its transpose. Hence, by using these we will prove $ AB = BA $ .
Formula used:
If $ A $ and $ B $ are symmetric matrix, then its transpose will be
 $ {\left( {AB} \right)^T} = AB $
And also,
 $ {B^T}{A^T} = {\left( {AB} \right)^T} $

Complete step-by-step answer:
 In the question, it is given that the $ A $ and $ B $ be two matrices which are symmetric so that means the order will be the same.
For proving this question, we will take two matrix
Let us assume, $ A = \left( {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right) $ and $ B = \left( {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right) $
Since its order is the same and also it is a square matrix therefore, we can say it is a symmetric matrix.
So from the formula, we have $ A = {A^T} $ and $ B = {B^T} $
Now we will calculate $ AB $ and $ BA $ .
Therefore, $ AB = \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right] $
So on applying the matrix multiplication, we have the matrices as
 $ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {0 \times 2 + 1 \times 1}&{0 \times 1 + 1 \times 2} \\
  {1 \times 2 + 0 \times 1}&{1 \times 1 + 0 \times 2}
\end{array}} \right] $
And on solving the multiplication of the above matrices, we get
\[ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&1
\end{array}} \right]\]
Similarly, we will find out the value for $ BA $
Therefore, $ BA = \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right] $
So on applying the matrix multiplication, we have the matrices as
\[ \Rightarrow BA = \left[ {\begin{array}{*{20}{c}}
  {2 \times 0 + 1 \times 1}&{2 \times 1 + 1 \times 0} \\
  {1 \times 0 + 2 \times 1}&{1 \times 1 + 2 \times 0}
\end{array}} \right]\]
And on solving the multiplication of the above matrices, we get
\[ \Rightarrow BA = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&1
\end{array}} \right]\]
So from here, we can see that the value of $ AB $ and $ BA $ are the same.
Hence, it is proved that $ AB = BA $ then $ AB $ is a symmetric matrix, we conclude this as when we take the transpose of AB it will be the same as AB.
So, the correct answer is “ $ AB $ is a symmetric matrix”.

Note: So when we have a question like that then the main concept used in this question will be we have to remember or memorize that in this type of matrix which has the order two the diagonal elements are always the same. And by using this we can solve such types of questions.