[M] In Exercises 37–40, use a matrix program to find the eigenvalues of the matrix. Then use the method of Example 4 with a row reduction routine to produce a basis for each eigenspace.

39. \(\left( {\begin{array}{*{20}{c}}4&{ - 9}&{ - 7}&8&2\\{ - 7}&{ - 9}&0&7&{14}\\5&{10}&5&{ - 5}&{ - 10}\\{ - 2}&3&7&0&4\\{ - 3}&{ - 13}&{ - 7}&{10}&{11}\end{array}} \right)\)

Write the command given below to input the matrix

\({\rm{ > > A = }}\left( {{\rm{4 - 9 - 7}}\,{\rm{8}}\,{\rm{2 ; - 7 - 9 0 }}\,{\rm{7 14; 5 10 5 - 5 - 10; - 2 3 7 0 4; - 3 - 13 - 7 10 11}}} \right)\)

\( > > {\rm{ ev}} = {\rm{eig}}\left( {\rm{A}} \right)\)

The output will be\({\rm{ev}} = \left( { - 2, - 2,5,5,5} \right)\).

Hence, the eigenvalues are \({\rm{ev}} = \left( { - 2, - 2,5,5,5} \right)\).

Write the command given below to find the null basis corresponding to \(\lambda = - 2\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 1 \right)*{\rm{eye}}\left( 5 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}{ - 2}&3\\7&7\\{ - 5}&{ - 5}\\5&0\\0&5\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = - 2\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\7\\{ - 5}\\5\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}3\\7\\{ - 5}\\0\\5\end{array}} \right)} \right\}\).

Write the command given below to find the null basis corresponding to \(\lambda = 5\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 2 \right)*{\rm{eye}}\left( 5 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}2&{ - 1}&2\\{ - 1}&1&0\\1&0&0\\0&1&0\\0&0&1\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = 5\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}2\\{ - 1}\\1\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\1\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\0\\0\\0\\1\end{array}} \right)} \right\}\).