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Mathematics LibreTexts

Section 10.3 Answers

  • Page ID
    30066
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    2. \({\bf y'}=\left[\begin{array}{cc}{0}&{1}\\{-\frac{P_{2}(x)}{P_{0}(x)}}&{-\frac{P_{1}(x)}{P_{0}(x)}}\end{array}\right]{\bf y}\)

    3. \({\bf y'}=\left[\begin{array}{cccc}{0}&{1}&{\ldots }&{0}\\{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{0}&{0}&{\ldots}&{1}\\{-\frac{P_{n}(x)}{P_{0}(x)}}&{-\frac{P_{n-1}(x)}{P_{0}(x)}}&{\ldots}&{-\frac{P_{1}(x)}{P_{0}(x)}}\end{array}\right]{\bf y}\)

    7. 

    b. \({\bf y}=\left[\begin{array}{c}{3e^{6t}-6e^{-2t}}\\{3e^{6t}+6e^{-2t}}\end{array}\right]\)

    c. \({\bf y}=\frac{1}{2}\left[\begin{array}{cc}{e^{6t}+e^{-2t}}&{e^{6t}-e^{-2t}}\\{e^{6t}-e^{-2t}}&{e^{6t}+e^{-2t}}\end{array}\right]{\bf k}\)

    8. 

    b. \({\bf y}=\left[\begin{array}{c}{6e^{-4t}+4e^{3t}}\\{6e^{-4t}-10e^{3t}}\end{array}\right]\)

    c. \({\bf y}=\frac{1}{7}\left[\begin{array}{cc}{5e^{-4t}+2e^{3t}}&{2e^{-4t}-2e^{3t}}\\{5e^{-4t}-5e^{3t}}&{2e^{-4t}+5e^{3t}}\end{array}\right]{\bf k}\)

    9. 

    b. \({\bf y}=\left[\begin{array}{c}{-15e^{2t}-4e^{t}}\\{9e^{2t}+2e^{t}}\end{array}\right]\)

    c. \({\bf y}=\left[\begin{array}{cc}{-5e^{2t}+6e^{t}}&{-10e^{2t}+10e^{t}}\\{3e^{2t}-3e^{t}}&{6e^{2t}-5e^{t}}\end{array}\right]{\bf k}\)

    10. 

    b. \({\bf y}=\left[\begin{array}{c}{5e^{3t}-3e^{t}}\\{5e^{3t}+3e^{t}}\end{array}\right]\)

    c. \({\bf y}=\frac{1}{2}\left[\begin{array}{cc}{e^{3t}+e^{t}}&{e^{3t}-e^{t}}\\{e^{3t}-e^{t}}&{e^{3t}+e^{t}}\end{array}\right]{\bf k}\)

    11. 

    b. \({\bf y}=\left[\begin{array}{c}{e^{2t}-2e^{3t}+3e^{-t}}\\{2e^{3t}-9e^{-t}}\\{e^{2t}-2e^{3t}+21e^{-t}}\end{array}\right]\)

    c. \({\bf y}=\frac{1}{6}\left[\begin{array}{ccc}{4e^{2t}+3e^{3t}-e^{-t}}&{6e^{2t}-6e^{3t}}&{2e^{2t}-3e^{3t}+e^{-t}}\\{-3e^{3t}+3e^{-t}}&{6e^{3t}}&{3e^{3t}-3e^{-t}}\\{4e^{2t}+3e^{3t}-7e^{-t}}&{6e^{2t}-6e^{3t}}&{2e^{2t}-3e^{3t}+7e^{-t}}\end{array}\right]{\bf k}\)

    12. 

    b. \({\bf y}=\frac{1}{3}\left[\begin{array}{c}{-e^{-2t}+e^{4t}}\\{-10e^{-2t}+e^{4t}}\\{11e^{-2t}+e^{4t}}\end{array}\right]\)

    c. \({\bf y}=\frac{1}{3}\left[\begin{array}{ccc}{2e^{-2t}+e^{4t}}&{-e^{-2t}+e^{4t}}&{-e^{-2t}+e^{4t}}\\{-e^{-2t}+e^{4t}}&{2e^{-2t}+e^{4t}}&{-e^{-2t}+e^{4t}}\\{-e^{-2t}+e^{4t}}&{-e^{-2t}+e^{4t}}&{2e^{-2t}+e^{4t}}\end{array}\right]{\bf k}\)

    13.

    b. \({\bf y}=\left[\begin{array}{c}{3e^{t}+3e^{-t}-e^{-2t}}\\{3e^{t}+2e^{-2t}}\\{-e^{-2t}}\end{array}\right]\)

    c. \({\bf y}=\left[\begin{array}{ccc}{e^{-t}}&{e^{t}-e^{-t}}&{2e^{t}-3e^{-t}+e^{-2t}}\\{0}&{e^{t}}&{2e^{t}-2e^{-2t}}\\{0}&{0}&{e^{-2}}\end{array}\right]{\bf k}\)

    14. \(YZ^{-1}\) and \(ZY^{-1}\)