Question

The inverse of symmetric matrix is
A. Symmetric
B. Skew-Symmetric
C. Diagonal matrix
D. None of these

Answer 1

Hint: First of all, consider a symmetric matrix of order $n$. Use the properties of transpose of the matrix to get the suitable answer for the given problem.

Complete step-by-step answer:
Let $A$ be an invertible symmetric matrix of order $n$.
$\therefore A{A^{ - 1}} = {A^{ - 1}}A = {I_n}..................................................\left( 1 \right)$
Now taking transpose on both sides we have,
$\Rightarrow {\left( {A{A^{ - 1}}} \right)^\prime } = {\left( {{A^{ - 1}}A} \right)^\prime } = {\left( {{I_n}} \right)^\prime }$
By using the formula ${\left( {AB} \right)^\prime } = B'A'$, we have
$\Rightarrow {\left( {{A^{ - 1}}} \right)^\prime }A' = A'{\left( {{A^{ - 1}}} \right)^\prime } = {\left( {{I_n}} \right)^\prime }$
As $A$ is a symmetric matrix $A' = A$ and for the identity matrix ${\left( {{I_n}} \right)^\prime } = {I_n}$

$$\Rightarrow {\left( {{A^{ - 1}}} \right)^\prime }A = A{\left( {{A^{ - 1}}} \right)^\prime } = {I_n} \\\ \therefore {\left( {{A^{ - 1}}} \right)^\prime } = A'{\text{ }}\left[ {{\text{Using }}\left( 1 \right)} \right] \\\$$

As the inverse of the matrix is unique ${A^{ - 1}}$ is symmetric.
Therefore, the inverse of a symmetric matrix is a symmetric matrix.
Thus, the correct option is A. a symmetric matrix

Note: A symmetric matrix is a square matrix that is equal to its transpose. ${A^{ - 1}}$ exists and is symmetric if and only if $A$ is symmetric. Remember this question as a statement for further references.