Answer:
Let A be a symmetric matrix and \[n\in N.\] Then, \[{{A}^{n}}=AAA...A\] upto n-times \[\Rightarrow \] \[{{({{A}^{n}})}^{T}}={{(AAA...A\,\,\text{upto}\,\,n\text{-times)}}^{T}}\] \[\Rightarrow \] \[{{({{A}^{n}})}^{T}}=({{A}^{T}}{{A}^{T}}{{A}^{T}}...{{A}^{T}}\,\,\text{upto}\,\,n\text{-times)}\] \[\Rightarrow \] \[{{({{A}^{n}})}^{T}}={{({{A}^{T}})}^{n}}={{A}^{n}}\] \[[\because \,\,{{A}^{T}}=A]\] Hence, \[{{A}^{n}}\] is also a symmetric matrix.
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