Math Insight

Review problems for exam 1

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  1. Find the eigenvalues of the matrix $$A=\left[\begin{matrix}0 & 1\\-4 & -4\end{matrix}\right]$$ Enter the eigenvalues separated by commas:

  2. Compute the following matrix-vector products.
    1. $\left[\begin{matrix}-4 & -5\\-6 & -2\end{matrix}\right]\left[\begin{matrix}1\\3\end{matrix}\right]=$
    2. $\left[\begin{matrix}5 & 5\\2 & 0\end{matrix}\right]\left[\begin{matrix}6\\5\end{matrix}\right]=$

  3. Find the determinant of the matrix $$\left[\begin{matrix}-1 & -1\\3 & -2\end{matrix}\right]$$

  4. Wegwarte
    Common chicory is perennial (it can live for several years) and typically produces bright blue flowers to reproduce. You decide to build a matrix model to describe a field of chicory where you track the amount of plant biomass in the non-flowering (stem, leaves, roots) and flowering parts of the plant. Assume that from each gram of non-flowering biomass, you get 0.8 grams of non-flowering biomass and 0.2 grams of flowering biomass the following month. From each gram of flowering biomass you get 4 grams of non-flowering biomass (reproduction) and 0.05 grams of flowering biomass.
    1. Write down a matrix model that describes this population.
    2. Assuming you start with 90 grams of non-flowering biomass and 3 grams of flowering biomass. How much biomass of each type will you have the following month? (Show your work with matrix vector multiplication.)
    3. Suppose you wanted to know how non-flowering biomass and flowering biomass you had the previous month. Write down the system of equations that you would use to solve this problem and solve this set of equations for the biomasses from the previous month.

      Hint: the matrix equation will look something like $A\vc{x} = \vc{b}$, where $\vc{b} = \begin{bmatrix} 90\\3\end{bmatrix}$.

    4. Find the eigenvalues of your matrix from part (a).
    5. Based on your answer to (c), what can you conclude about the long-term growth of this chicory population?

  5. Consider the following $2$-dimensional system of discrete dynamical equations: \begin{eqnarray*} x_{n+1} &=& 5 x_{n} + 3 y_{n}\\ y_{n+1} &=& - 18 x_{n} - 10 y_{n}\\ x_{0} &=& -10\\ y_{0} &=& -2 \end{eqnarray*}
    1. Compute the next five values of $x$ and $y$.
      $x_1 =$
      ,  $y_1 =$

      $x_2 =$
      ,  $y_2 =$

      $x_3 =$
      ,  $y_3 =$

      $x_4 =$
      ,  $y_4 =$

      $x_5 =$
      ,  $y_5 =$
    2. Convert the system into a matrix equation.

      $\begin{bmatrix} x_{n+1}\\ y_{n+1} \end{bmatrix}=$




      $\begin{bmatrix} x_n\\ y_n\end{bmatrix}$
    3. Find the eigenvalues and eigenvectors of the matrix you found in part a. Enter the eigenvalues separated by commas, in increasing order:

      Enter the eigenvectors in the same order as the corresponding eigenvalues:
      ,
    4. Which eigenvector direction will $(x_n, y_n)$ approach as $n$ goes to infinity?

  6. Find all solutions to the following system of equations, or explain why none exist. \begin{eqnarray*} 5 x + 3 y&=&-8\\ 15 x + 9 y&=&-24 \end{eqnarray*}

  7. Consider the dynamical system $\vc{x}_{n+1} = A\vc{x}_n$, where $A$ is a $3 \times 3$ matrix. Use the following information about the behavior of the dynamical system to estimate the dominant eigenvalue of $A$ and its eigenvector.

    For the initial condition $\vc{x}_0 = \left[\begin{matrix}0.1\\0.5\\1.6\end{matrix}\right]$, iterating the model forward 10 times gives: \begin{gather*}\vc{x}_1 = \left[\begin{matrix}1.71\\1.61\\0.24\end{matrix}\right], \vc{x}_2 = \left[\begin{matrix}3.445\\1.041\\0.698\end{matrix}\right], \vc{x}_3 = \left[\begin{matrix}6.693\\1.63\\1.207\end{matrix}\right], \vc{x}_4 = \left[\begin{matrix}12.79\\2.914\\2.292\end{matrix}\right], \vc{x}_5 = \left[\begin{matrix}24.39\\5.495\\4.358\end{matrix}\right]\\ \vc{x}_6 = \left[\begin{matrix}46.48\\10.45\\8.302\end{matrix}\right], \vc{x}_7 = \left[\begin{matrix}88.58\\19.9\\15.82\end{matrix}\right], \vc{x}_8 = \left[\begin{matrix}168.8\\37.92\\30.15\end{matrix}\right], \vc{x}_9 = \left[\begin{matrix}321.7\\72.27\\57.45\end{matrix}\right], \vc{x}_{10} = \left[\begin{matrix}613.0\\137.7\\109.5\end{matrix}\right]\end{gather*}

    For the initial condition $\vc{x}_0 = \left[\begin{matrix}0.8\\1.3\\1.7\end{matrix}\right]$, iterating the model forward 10 times gives: \begin{gather*}\vc{x}_1 = \left[\begin{matrix}3.15\\2.08\\0.54\end{matrix}\right], \vc{x}_2 = \left[\begin{matrix}6.257\\1.74\\1.207\end{matrix}\right], \vc{x}_3 = \left[\begin{matrix}12.07\\2.86\\2.172\end{matrix}\right], \vc{x}_4 = \left[\begin{matrix}23.05\\5.227\\4.125\end{matrix}\right], \vc{x}_5 = \left[\begin{matrix}43.94\\9.89\\7.849\end{matrix}\right]\\ \vc{x}_6 = \left[\begin{matrix}83.74\\18.82\\14.96\end{matrix}\right], \vc{x}_7 = \left[\begin{matrix}159.6\\35.85\\28.5\end{matrix}\right], \vc{x}_8 = \left[\begin{matrix}304.1\\68.32\\54.31\end{matrix}\right], \vc{x}_9 = \left[\begin{matrix}579.5\\130.2\\103.5\end{matrix}\right], \vc{x}_{10} = \left[\begin{matrix}1104.0\\248.1\\197.2\end{matrix}\right]\end{gather*}

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Math 2241, Spring 2023