Chapter 9: Problem 1
\(\mathbf{A}=\left[ \begin{array}{ll}{2} & {-4} \\ {2} & {-2}\end{array}\right]\)
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Chapter 9: Problem 1
\(\mathbf{A}=\left[ \begin{array}{ll}{2} & {-4} \\ {2} & {-2}\end{array}\right]\)
These are the key concepts you need to understand to accurately answer the question.
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\(\mathbf{A}=\left[ \begin{array}{lll}{0} & {2} & {2} \\ {2} & {0} & {2} \\\ {2} & {2} & {0}\end{array}\right]\)
\(\mathbf{A}=\left[ \begin{array}{rr}{0} & {-1} \\ {1} & {0}\end{array}\right], \quad \mathbf{f}(t)=\left[ \begin{array}{l}{t^{2}} \\\ {1}\end{array}\right]\)
(a) Show that the matrix \(\mathbf{A}=\left[ \begin{array}{lll}{2} & {1} & {6} \\ {0} & {2} & {5} \\\ {0} & {0} & {2}\end{array}\right]\) has the repeated eigenvalue \(r=2\) with multiplicity 3 and that all the eigenvectors of \(\mathbf{A}\) are of the form \(\mathbf{u}=s \operatorname{col}(1,0,0)\) (b) Use the result of part (a) to obtain a solution to the system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\) of the form \(\mathbf{x}_{1}(t)=e^{2 t} \mathbf{u}_{1} .\) (c) To obtain a second linearly independent solution to \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},\) try \(\mathbf{x}_{2}(t)=t e^{2 t} \mathbf{u}_{1}+e^{2 t} \mathbf{u}_{2} .[\)Hint\(:\) Show that \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) must satisfy \((\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{1}=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{2}=\mathbf{u}_{1} \cdot ]\) (d) To obtain a third linearly independent solution to \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},\) try \(\quad \mathbf{x}_{3}(t)=\frac{t^{2}}{2} e^{2 t} \mathbf{u}_{1}+t e^{2 t} \mathbf{u}_{2}+e^{2 t} \mathbf{u}_{3}\) [Hint: Show that \(\mathbf{u}_{1}, \mathbf{u}_{2},\) and \(\mathbf{u}_{3}\) must satisfy \((\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{1}=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{2}=\mathbf{u}_{1}\) \((\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{3}=\mathbf{u}_{2} . ]\) (e) Show that \((\mathbf{A}-2 \mathbf{I})^{2} \mathbf{u}_{2}=(\mathbf{A}-2 \mathbf{I})^{3} \mathbf{u}_{3}=\mathbf{0}\)
\(\mathbf{A}=\left| \begin{array}{rrrrr}{0} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {-1} & {-3} & {-3} & {0} & {0} \\ {0} & {0} & {0} & {0} & {1} \\ {0} & {0} & {0} & {-4} & {-4}\end{array}\right|\)
\(\mathbf{X}(t)=\left[ \begin{array}{ccc}{e^{t}} & {e^{-t}} & {e^{2 t}} \\\ {e^{t}} & {-e^{-t}} & {2 e^{2 t}} \\ {e^{t}} & {e^{-t}} & {4 e^{2 t}}\end{array}\right]\)
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