Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

Eigenvalue:An eigenvector of \(n \times n\) the matrix \(A\) is a nonzero vector \({\bf{x}}\) such that \(A{\bf{x}} = \lambda {\bf{x}}\) for some scalar \(\lambda \) where scalar \(\lambda \) is called an eigenvalue of \(A\).

Assume\(A = \left( {\begin{aligned}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{aligned}} \right)\).

From the exercise\(15\), the eigenvalues are \(\lambda = a - b\) and \(\lambda = a + \left( {n - 1} \right)b\).

Therefore,

Put \(a = 1\), \(b = 2\) and \(n = 3\).

\(\begin{aligned}{c}\lambda &= a - b\\ &= 1 - 2\\ &= - 1\end{aligned}\)

\(\begin{aligned}{c}\lambda &= a + \left( {n - 1} \right)b\\ &= 1 + \left( {3 - 1} \right)2\\ &= 5\end{aligned}\)

Thus, eigenvalues for the matrix \(\left( {\begin{aligned}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{aligned}} \right)\) are \( - 1\) and \(5\).

Assume\(A = \left( {\begin{aligned}{*{20}{c}}7&3&3&3&3\\3&7&3&3&3\\3&3&7&3&3\\3&3&3&7&3\\3&3&3&3&7\end{aligned}} \right)\).

From the exercise \(15\), the eigenvalues are \(\lambda = a - b\) and \(\lambda = a + \left( {n - 1} \right)b\).

Therefore,

Put \(a = 7\), \(b = 3\) and \(n = 5\).

\(\begin{aligned}{c}\lambda &= a - b\\ &= 7 - 3\\ &= 5\end{aligned}\)

\(\begin{aligned}{c}\lambda &= a + \left( {n - 1} \right)b\\ &= 7 + \left( {5 - 1} \right)3\\ &= 19\end{aligned}\)

Thus, the eigenvalues for the matrix \(\left( {\begin{aligned}{*{20}{c}}7&3&3&3&3\\3&7&3&3&3\\3&3&7&3&3\\3&3&3&7&3\\3&3&3&3&7\end{aligned}} \right)\) are \(5\) and \(19\).