Given the following system $ \ \vec{x'}=\begin{pmatrix}12 & 25 \\ 1 & 225 \end{pmatrix} \vec{x} $ .
Find the general solution by eigen value method .
Answer:
The coefficient matrix has the Eigen values $ \ 11.9 \ \ and \ \ 225.1 \ $ .
But I can not find the Eigen vectors from these Eigen values because substituting each Eigen value in the system
$ \begin{pmatrix}12-\lambda & 25 \\ 1 & 225-\lambda \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=0 $ ,
it produces two different homogeneous equations giving $ \ \vec{0} $ .
Thus how can I find the Eigen vectors ? Any help ?
Eigenvalues are $$\lambda_1=\frac{1}{2}(237+\sqrt{45469})$$ $$\lambda_1=\frac{1}{2}(237-\sqrt{45469})$$ Substituting into $$(A-I\lambda)x=0$$ yields a pair of eigenvectors $$x_1=\left(\frac{1}{2}(-213+\sqrt{45469},1\right)$$ $$x_2=\left(\frac{1}{2}(-213-\sqrt{45469}),1\right)$$