Question

If A and B are symmetric matrices of the same order, Prove that AB – BA is a symmetric matrix.

Answer 1

Hint – In this particular question use the concept that a symmetric matrix is always a square matrix (i.e. rows and columns are the same) and it is always equal to its transpose so use these concepts to reach the solution of the question.

Complete step-by-step answer:
As we all know a symmetric matrix is a square matrix which is always equal to its transpose of the matrix.
For example:
Consider a matrix M = \[\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]\]
Now take the transpose of this matrix we have,
$ \Rightarrow {M^T} = {\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]^T}$, where T is the sign for the transpose.
Now apply the transpose (i.e. in transpose rows changed into column and column changed into rows).
$ \Rightarrow {M^T} = \left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]$
Therefore, $M = {M^T}$
So as we see that M and ${M^T}$ are the same therefore it is called a symmetric matrix.
Now it is given that A and B are the symmetric matrix of the same order.
So consider any symmetric matrix of order $\left( {2 \times 2} \right)$
Let, A = \[\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]\] and B = \[\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right]\]
So it is a symmetric matrix
Therefore, $A = {A^T}{\text{ and }}B = {B^T}$
Now we have to prove AB – BA is a symmetric matrix.
Proof –
Now first find out the values of AB and BA
Therefore, AB = \[\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right]\]
 Now apply matrix multiplication we have,
$ \Rightarrow $AB = \[\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {0 \times 2 + 1 \times 1}&{0 \times 1 + 1 \times 2} \\
  {1 \times 2 + 0 \times 1}&{1 \times 1 + 0 \times 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&1
\end{array}} \right]\]
Now find out BA
Therefore, BA = \[\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  1&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {2 \times 0 + 1 \times 1}&{2 \times 1 + 1 \times 0} \\
  {1 \times 0 + 2 \times 1}&{1 \times 1 + 2 \times 0}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&1
\end{array}} \right]\]
So as we see that the value of the AB and BA are coming the same.
So the difference of these matrix is a zero matrix which is given as,
Therefore, AB – BA = \[\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  2&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]\]
Now take the transpose of the above matrix we have,
\[ \Rightarrow {\left( {AB - BA} \right)^T} = {\left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right] = \] (AB – BA)
So the condition of symmetricity is satisfied.
Hence proved.

Note – Whenever we face such types of question the key concept we have to remember is that in a symmetric matrix (consider an order ($2 \times 2$)), the diagonal elements are always equal, so first assume any two symmetric matrix A and B as above, then calculate the values of AB and BA as above, then calculate (AB – BA) and its transpose we get the required result.