EE.C10 - Find the characteristic polynomial of the Matrix
\[
A =
\left(\begin{array}{ccc}
1 & 2\\
3 & 4 \\
\end{array}\right)
\] Since it has already been done; I’ll make a new matrix to solve. Same parameters. To my understanding; to do this you multiply the matrix by lambda times the identity matrix; then take the determinant of your result to find the characteristic polynomial.
Lets try it
\[ B = \left(\begin{array}{ccc} 4 & 3\\ 2 & 1 \\ \end{array}\right) \] \[ B = \left(\begin{array}{ccc} 4 & 3\\ 2 & 1 \\ \end{array}\right) - \lambda\left(\begin{array}{ccc} 1 & 0\\ 0 & 1 \\ \end{array}\right)= \left(\begin{array}{ccc} 4-\lambda & 3\\ 2 & 1-\lambda \\ \end{array}\right) \] \[ det\left(\begin{array}{ccc} 4-\lambda & 3\\ 2 & 1-\lambda \\ \end{array}\right)= (\lambda-5)\lambda - 2 \]
The characteristic polynomial of matrix B is
\[
(\lambda-5)\lambda - 2
\]
Below is the start of EE.C22; will update tomorrow with solution. Estimating quite a bit of lateX formatting.
\[ B = \left(\begin{array}{ccc} 2 & -1\\ 1 & 1 \\ \end{array}\right) - \left(\begin{array}{ccc} \lambda & 0\\ 0 & \lambda \\ \end{array}\right) = |A-\lambda\cdot I| \]