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

3. \(\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]\)

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 \(A = \left[ {\begin{array}{*{20}{c}}3&- 2\\1&- 1\end{array}} \right]\) 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:

\[\begin{array} - \lambda I = \left[ {\begin{array}{*{20}{c}}3&- 2\\1&- 1\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right]\end{array}\]

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

\[\begin{array}det\left( {A - \lambda I} \right) = det\left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\left( { - 1 - \lambda } \right) + 2\\ = {\lambda ^2} - 2\lambda - 1\end{array}\]

So, the characteristic polynomial of is \({\lambda ^2} - 2\lambda - 1\).

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\(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right] = 0\\\left( {3 - \lambda } \right)\left( { - 1 - \lambda } \right) + 2 = 0\\{\lambda ^2} - 2\lambda - 3 + 2 = 0\\{\lambda ^2} - 2\lambda - 1 = 0\end{array}\]

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\[{\lambda ^2} - 2\lambda - 1 = 0\]is obtained as follows:

\[\begin{array}{\lambda ^2} - 2\lambda - 1 = 0\\\lambda = \frac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( { - 1} \right)} }}{2}\\ = \frac{{2 \pm 2\sqrt 2 }}{2}\\ = 1 \pm \sqrt 2 \end{array}\]

The eigenvalues of \(A\) are \(\lambda = 1 + \sqrt 2 \) and \(\lambda = 1 - \sqrt 2 \).