Suppose that the matrix $\mathrm{A}$ has the following eigenvalues and eigenvectors:
    $\lambda_1 = 4$ with $\displaystyle\boldsymbol{\vec{v}_1} = \left\lbrack \begin{array}{r} -3 \\ 1 \end{array} \right\rbrack.$
and
    $\lambda_2 = 0$ with $\displaystyle\boldsymbol{\vec{v}_2} = \left\lbrack \begin{array}{r} -3 \\ 3 \end{array} \right\rbrack.$

Write the solution to the linear system $\boldsymbol{\vec{r}\,'} = \mathrm{A}\boldsymbol{\vec{r}}$ in the following forms.

A. In eigenvalue/eigenvector form:

$\qquad \displaystyle \biggl\lbrack\begin{array}{c} x(t) \\ y(t)\end{array}\biggr\rbrack = c_1$

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

$\displaystyle e$   $t$

$\ + \ c_2$

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

$\displaystyle e$   $t$

B. In fundamental matrix form:

$\qquad \displaystyle \biggl\lbrack\begin{array}{c} x(t) \\ y(t)\end{array}\biggr\rbrack =$

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

$\displaystyle \left\lbrack \begin{array}{r} c_1 \\ c_2 \end{array} \right\rbrack$

C. As two equations: (write "c1" and "c2" for $c_1$ and $c_2$ )

$\qquad x(t) =$

$\qquad y(t) =$

Note: if you are feeling adventurous you could use other eigenvectors like $4\,\boldsymbol{\vec{v}_1}$ or $-3\,\boldsymbol{\vec{v}_2}$.