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.