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 = $$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$
Now take the transpose of this matrix we have,
$\Rightarrow M^T={\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^T$, 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).
$\Rightarrow M^T=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
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 = $$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$ and B = $$\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]$$
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 = $$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]$$
 Now apply matrix multiplication we have,
$\Rightarrow$AB = $$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0\times 2+1\times 1& 0\times 1+1\times 2\\ 1\times 2+0\times 1& 1\times 1+0\times 2\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$$
Now find out BA
Therefore, BA = $$\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}2\times 0+1\times 1& 2\times 1+1\times 0\\ 1\times 0+2\times 1& 1\times 1+2\times 0\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$$
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 = $$\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$
Now take the transpose of the above matrix we have,
$$\Rightarrow {\left(AB-BA\right)}^T={\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]}^T=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=$$ (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.