Question

If A and B are symmetric matrices and AB = BA then ${{A}^{-1}}B$ is a
(a) Symmetric matrix
(b) Skew symmetric matrix
(c) Identity matrix
(d) None of these

Answer 1

To solve this question we first need to know when is a matrix A called symmetric. It is so when, ${{A}^{'}}=A$. Now to compute if ${{A}^{-1}}B$ is symmetric or skew symmetric compute the transpose of ${{A}^{-1}}B$. Use the given theory AB = BA to conclude the answer. Remember that a matrix is skew symmetric if ${{A}^{'}}=-A$.

Complete step-by-step answer :
A matrix is said to be symmetric if its transpose is equal to the matrix itself; i.e. M’ = M, where M is matrix. Given that A and B are symmetric matrices; which means that, A’ = A and B’ = B.
Also we are given that, AB = BA.
We have;
AB = BA – (1)
Now as B’ = B & A’ = A, then substituting this in Right hand side of (1) we get;
AB = B’ A’
$\Rightarrow$ AB = B’ A’ = (AB)’
So we get;
AB = (AB)’
Hence AB is symmetric.
Now consider $AB{{A}^{-1}}$,
Now as AB = BA
$\Rightarrow AB{{A}^{-1}}=BA{{A}^{-1}}$
And $A{{A}^{-1}}$ = Identity matrix
$\Rightarrow AB{{A}^{-1}}=B$
Applying ${{A}^{-1}}$ on both sides of above equation we get;
$\Rightarrow {{A}^{-1}}AB{{A}^{-1}}={{A}^{-1}}B$
$\Rightarrow B{{A}^{-1}}={{A}^{-1}}B$ - (2)
Finally consider (${{A}^{-1}}B$)’
Using equation (2), we have,

& \Rightarrow \left( {{A}^{-1}}B \right)'=\left( B{{A}^{-1}} \right)' \\\ & \Rightarrow \left( {{A}^{-1}}B \right)'=\left( {{A}^{-1}} \right)'B' \\\ \end{aligned}$$ Now because A is symmetric, then $${{A}^{-1}}$$ is also symmetric. $$\Rightarrow \left( {{A}^{-1}} \right)'={{A}^{-1}}$$ Using this in above equation we get, $$\Rightarrow \left( {{A}^{-1}}B \right)'=\left( {{A}^{-1}} \right)'B'={{A}^{-1}}B'$$ Now again as B is symmetric $$\Rightarrow B'=B$$. Then, $$\left( {{A}^{-1}}B \right)'=\left( {{A}^{-1}} \right)'B'={{A}^{-1}}B'$$. Now again as B is symmetric $$\Rightarrow B'=B$$. Then, $$\left( {{A}^{-1}}B \right)'={{A}^{-1}}B'={{A}^{-1}}B$$. Hence we get, $$\left( {{A}^{-1}}B \right)'={{A}^{-1}}B$$. Hence, $${{A}^{-1}}B$$ is a symmetric matrix, option (a) is correct. **Note** : The possibility of mistake in this question can be where the matrix $${{A}^{-1}}$$ is assumed to be symmetric because A is so. If A is symmetric then, $$A'=A$$. And $$\left( {{A}^{-1}} \right)'={{\left( A' \right)}^{-1}}={{A}^{-1}}$$ as $$A'=A$$. Therefore, when A is a symmetric matrix then $${{A}^{-1}}$$ is also a symmetric matrix. Hence we can use this easily.