Differential Equations and Linear Algebra, 4th Edition

Section 6.2

1. 1. ${\lambda }_1=1,{\lambda }_2=3;\mathbf{P}=\left[\begin{array}{rr}1& 2\\ 1& 1\end{array}\right],\mathbf{D}=\left[\begin{array}{ll}1& 0\\ 0& 3\end{array}\right]$

2. 2. ${\lambda }_1=0,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}1& 3\\ 1& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{ll}0& 0\\ 0& 2\end{array}\right]$

3. 3. ${\lambda }_1=2,{\lambda }_2=3;\mathbf{P}=\left[\begin{array}{ll}1& 3\\ 1& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{ll}2& 0\\ 0& 3\end{array}\right]$

4. 4. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}1& 4\\ 1& 3\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

5. 5. ${\lambda }_1=1,{\lambda }_2=3;\mathbf{P}=\left[\begin{array}{rr}1& 4\\ 1& 3\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 3\end{array}\right]$

6. 6. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}2& 3\\ 3& 4\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

7. 7. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}2& 5\\ 1& 2\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

8. 8. ${\lambda }_1=1,{\lambda }_2=2;\mathbf{P}=\left[\begin{array}{rr}3& 5\\ 2& 3\end{array}\right],\mathbf{D}=\left[\begin{array}{rr}1& 0\\ 0& 2\end{array}\right]$

9. 9. The double eigenvalue ${\lambda }_1={\lambda }_2=1$ has only the single associated eigenvector ${\mathbf{v}}_1=(2,1)$, so the matrix A is not diagonalizable.

10. 10. The double eigenvalue ${\lambda }_1={\lambda }_2=2$ has only the single associated eigenvector ${\mathbf{v}}_1=(1,1)$, so the matrix A is not diagonalizable.

11. 11. The double eigenvalue ${\lambda }_1={\lambda }_2=2$ has only the single associated eigenvector ${\mathbf{v}}_1=(-1,3)$, so the matrix A is not diagonalizable.

12. 12. The double eigenvalue ${\lambda }_1={\lambda }_2=-1$ has only the single associated eigenvector ${\mathbf{v}}_1=(-$