Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

5. \(\left[ {\begin{array}{*{20}{c}}2&1\\-1&4\end{array}} \right]\)

Ifis an\(n \times n\)matrix, then\(det\left( {A - \lambda I} \right)\), which is a polynomial of degree\(n\), is called the characteristic polynomial of\(A\).

It is given that \(A = \left[ {\begin{array}{*{20}{c}}2&1\\ - 1&4\end{array}} \right]\) and \(I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}A - \lambda I = \left[ {\begin{array}{*{20}{c}}2&1\\ - 1&4\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{2 - \lambda }&1\\{ - 1}&{4 - \lambda }\end{array}} \right]\end{array}\]

Now, calculate the determinant of the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\[\begin{array}det\left( {A - \lambda I} \right) = det\left[ {\begin{array}{*{20}{c}}{2 - \lambda }&1\\{ - 1}&{4 - \lambda }\end{array}} \right]\\ = \left( {2 - \lambda } \right)\left( {4 - {\rm{\lambda }}} \right) + 1\\ = {\lambda ^2} - 6\lambda + 8 + 1\\ = {\lambda ^2} - 6\lambda + 9\end{array}\]

So, the characteristic polynomial of is \[{\lambda ^2} - 6\lambda + 9\].

To find the eigenvalues of the matrix, we must calculate all the scalarssuch that\(\left( {A - \lambda I} \right)x = 0\) has a non-trivial solution which is equivalent to finding allsuch that the matrix\(\left( {A - \lambda I} \right)\)is not invertible, that is, when determinant of\(\left( {A - \lambda I} \right)\)is zero.

Thus, the eigenvalues of \(A\) are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, find the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}{2 - \lambda }&1\\{ - 1}&{4 - \lambda }\end{array}} \right] = 0\\\left( {2 - \lambda } \right)\left( {4 - {\rm{\lambda }}} \right) + 1 = 0\\{\lambda ^2} - 6\lambda + 8 + 1 = 0\\{\lambda ^2} - 6\lambda + 9 = 0\end{array}\]

For the quadratic equation,\(a{x^2} + bx + c = 0\), the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation\[{\lambda ^2} - 6\lambda + 9 = 0\]is obtained as follows:

\[\begin{array}{\lambda ^2} - 6\lambda + 9 = 0\\\lambda = \frac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 9 \right)} }}{2}\\ = \frac{{6 \pm 0}}{2}\\ = 3,3\end{array}\]

The matrix \(A\) has single eigenvalue \(\lambda = 3\) with a multiplicity of 2.