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,

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.