Problem T10 page 402

Solution

\(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)

If \(\lambda\) is an eigen value of A the the characteristic polynomial is as follows:

\(det\begin{Bmatrix} \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 0 \end{Bmatrix}\)

\(det\begin{Bmatrix} \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} - \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 0 \end{Bmatrix}\)

\(det \begin{bmatrix} \lambda - a & - b \\ -c & \lambda - d \end{bmatrix} = 0\)

This simplifies to:

\((\lambda - a)(\lambda - d) - (-b)(-c)\)

\(\lambda^2 - \lambda a - \lambda d + ad - bc\)

The the constant term is ad - bc