In exercise [1, p. 34 exercise 2.7] it is requested to find the inverse of the real symmetric Toeplitz matrix
\mathbf{A} = \left[ {\begin{array}{*{20}c} 1 & { - a} & {a^2 } \\ { - a} & 1 & { - a} \\ {a^2 } & { - a} & 1 \\ \end{array}} \right] (1)

and show that it is symmetric and persymmetric.
Solution: The cofactors c_{ij} of the matrix \mathbf{A} are given by:
c_{11}=( - 1)^{(1 + 1)} \left| {\begin{array}{*{20}c} 1 & { - a} \\ { - a} & 1 \\ \end{array}} \right| = 1 - a^2
c_{12}=( - 1)^{(1 + 2)} \left| {\begin{array}{*{20}c} { - a} & { - a} \\ {a^2 } & 1 \\ \end{array}} \right| = a - a^3
c_{13}=( - 1)^{(1 + 3)} \left| {\begin{array}{*{20}c} { - a} & 1 \\ {a^2 } & {-a} \\ \end{array}} \right| = 0
c_{21}=( - 1)^{(2 + 1)} \left| {\begin{array}{*{20}c} { - a} & {a^2} \\ {-a } & {1} \\ \end{array}} \right| =a-a^3
c_{22}=( - 1)^{(2 + 2)} \left| {\begin{array}{*{20}c} { 1} & {a^2} \\ {a^2} & {1} \\ \end{array}} \right| =1-a^4
c_{23}=( - 1)^{(2 + 3)} \left| {\begin{array}{*{20}c} { 1} & {-a} \\ {a^2 } & {-a} \\ \end{array}} \right| =a-a^3
c_{31}=( - 1)^{(3 + 1)} \left| {\begin{array}{*{20}c} { -a} & {a^2} \\ {1} & {-a} \\ \end{array}} \right| =0
c_{32}=( - 1)^{(3 + 2)} \left| {\begin{array}{*{20}c} { 1} & {a^2} \\ {-a} & {-a} \\ \end{array}} \right| =a-a^3
c_{33}=( - 1)^{(3 + 3)} \left| {\begin{array}{*{20}c} { 1} & {-a} \\ {-a } & {1} \\ \end{array}} \right| =1-a^2

Thus the determinant of \mathbf{A} is given by (let i=1):
Det(\mathbf{A})=\sum\limits_{j=1}^{3}a_{ij}C_{ij}
=1\cdot (1-a^2)+ (-a)\cdot(a-a^3)+a^2 \cdot 0
=(1-a^2)^2

The inverse of the matrix \mathbf{A} is thus equal to:
\mathbf{A}^{-1}=\frac{1}{det{\mathbf{A}}}\left[ {\begin{array}{*{20}c} c_{11} & c_{21} & c_{31} \\ c_{12} & c_{22} & c_{32} \\ c_{13} & c_{23} & c_{33} \\ \end{array}} \right]
=\frac{1}{(1-a^2)^2}\left[ {\begin{array}{*{20}c} 1-a^2 & a-a^3 & 0 \\ a-a^3 & 1-a^4 & a-a^3 \\ 0 & a-a^3 & 1-a^2 \\ \end{array}} \right]

As it can be seen the matrix is persymmetric because the matrix elements [\mathbf{A}^{-1}]_{ij}=m_{ij} satisfy the relationship m_{ij}=m_{(4-i)(4-j)}. It is also symmetric because m_{ij} =m_{ji}. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.