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

Hint: 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,
\[\begin{align}
  & \Rightarrow \left( {{A}^{-1}}B \right)'=\left( B{{A}^{-1}} \right)' \\
 & \Rightarrow \left( {{A}^{-1}}B \right)'=\left( {{A}^{-1}} \right)'B' \\
\end{align}\]
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.