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Expert-verified Found in: Page 426 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Let be an matrix and a scalar. Consider the following two systems:$\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}\left(l\right)\phantom{\rule{0ex}{0ex}}\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{c}}}{\mathbf{dt}}{\mathbf{=}}\left(A+{\mathrm{kl}}_{n}\right)\stackrel{\mathbf{\to }}{\mathbf{c}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}\left(\mathrm{ll}\right)$Show that if $\stackrel{\mathbf{\to }}{\mathbf{x}}\left(t\right)$ is a solution of the system (l) then role="math" localid="1659701582223" $\stackrel{\mathbf{\to }}{\mathbf{c}}\left(t\right){\mathbf{=}}{{\mathbit{e}}}^{\mathbf{k}\mathbf{t}}\stackrel{\mathbf{\to }}{\mathbf{x}}\left(t\right)$ is a solution of the system (ll).

Yes, $\stackrel{\to }{c}\left(t\right)={e}^{kt}\stackrel{\to }{x}\left(t\right)$ is a solution of the system $\frac{\mathrm{d}\stackrel{\to }{\mathrm{c}}}{\mathrm{dt}}=\left(\mathrm{A}+{\mathrm{kl}}_{\mathrm{n}}\right)\stackrel{\to }{\mathrm{c}}$.

See the step by step solution

## Step 1: Determine the equation of the solution.

Consider $\stackrel{\to }{x}\left(t\right)$ and $\stackrel{\to }{c}\left(t\right)$ is a solution of the system $\frac{d\stackrel{\to }{x}}{dt}=A\stackrel{\to }{x}$ and $\frac{\mathrm{d}\stackrel{\to }{\mathrm{c}}}{\mathrm{dt}}=\left(\mathrm{A}+{\mathrm{kl}}_{\mathrm{n}}\right)\stackrel{\to }{\mathrm{c}}$ respectively.

Assume ${\lambda }_{1},{\lambda }_{2},{\lambda }_{3},\dots ,{\lambda }_{n}$ be the Eigen values of the matrix A then there exist Eigen vectors ${\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{3},\dots ,{\stackrel{\to }{v}}_{n}$ such that ${\stackrel{\to }{x}}_{1}\left(t\right)={c}_{1}{e}^{{\lambda }_{1}t}{\stackrel{\to }{v}}_{1}+{c}_{2}{e}^{{\lambda }_{2}t}{\stackrel{\to }{v}}_{2}+...+{c}_{n}{e}^{{\lambda }_{n}t}{\stackrel{\to }{v}}_{n}$ and $\stackrel{\to }{c}\left(t\right)={c}_{1}{e}^{\left({\lambda }_{1}+k\right)}{\stackrel{\to }{v}}_{1}+{c}_{2}{e}^{\left({\lambda }_{2}+k\right)}{\stackrel{\to }{v}}_{2}+...+{c}_{n}{e}^{\left({\lambda }_{n}+k\right)}{\stackrel{\to }{v}}_{n}$ where ${c}_{1},{c}_{2},\dots ,{c}_{n}$ is constant.

If x(t) is the solution of the linear system localid="1659702528208" $\frac{d\stackrel{\to }{x}}{dt}{=}{A}\stackrel{\to }{x}$ then ${\stackrel{\to }{x}}_{{1}}{\left(}{t}{\right)}{=}{{c}}_{{1}}{{e}}^{{\lambda }_{1}t}{\stackrel{\to }{v}}_{{1}}{+}{{c}}_{{2}}{{e}}^{{\lambda }_{2}t}{\stackrel{\to }{v}}_{{2}}{+}{}{.}{.}{.}{}{+}{{c}}_{{n}}{{e}}^{{\lambda }_{n}t}{\stackrel{\to }{v}}_{{n}}$ where ${{\lambda }}_{{1}}{,}{{\lambda }}_{{2}}{,}{{\lambda }}_{{3}}{,}{\dots }{,}{{\lambda }}_{{n}}$ be the Eigen values and of ${n}{×}{n}$ matrix A.

## Step 2: Show that c→(t)=ektx→(t) is a solution of the system.

Simplify the equation ${\stackrel{\to }{x}}_{1}\left(t\right)={c}_{1}{e}^{{\lambda }_{1}t}{\stackrel{\to }{v}}_{1}+{c}_{2}{e}^{{\lambda }_{2}t}{\stackrel{\to }{v}}_{2}+...+{c}_{n}{e}^{{\lambda }_{n}t}{\stackrel{\to }{v}}_{n}$ as follows.

$\stackrel{\to }{c}\left(t\right)={c}_{1}{e}^{\left({\lambda }_{1}+k\right)}{\stackrel{\to }{v}}_{1}+{c}_{2}{e}^{\left({\lambda }_{2}+k\right)}{\stackrel{\to }{v}}_{2}+...+{c}_{n}{e}^{\left({\lambda }_{n}+k\right)}{\stackrel{\to }{v}}_{n}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{c}\left(t\right)={c}_{1}{e}^{{\lambda }_{1}t}{e}^{kt}{\stackrel{\to }{v}}_{1}+{c}_{2}{e}^{{\lambda }_{2}t}{e}^{kt}{\stackrel{\to }{v}}_{2}+...+{c}_{n}{e}^{{\lambda }_{n}t}{e}^{kt}{\stackrel{\to }{v}}_{n}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{c}\left(t\right)={e}^{kt}\left\{{c}_{1}{e}^{{\lambda }_{1}t}{\stackrel{\to }{v}}_{1}+{c}_{2}{e}^{{\lambda }_{2}t}{\stackrel{\to }{v}}_{2}+...+{c}_{n}{e}^{{\lambda }_{n}t}{\stackrel{\to }{v}}_{n}\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{c}\left(t\right)={e}^{kt}\stackrel{\to }{x}\left(t\right)$

By the definition of solution of the linear system, $\stackrel{\to }{c}\left(t\right)={e}^{kt}\stackrel{\to }{x}\left(t\right)$ is a solution.

Hence, if $\stackrel{\to }{x}\left(t\right)$ and $\stackrel{\to }{c}\left(t\right)$ is a solution of the system $\frac{d\stackrel{\to }{x}}{dt}=A\stackrel{\to }{x}$ and $\frac{\mathrm{d}\stackrel{\to }{\mathrm{c}}}{\mathrm{dt}}=\left(\mathrm{A}+{\mathrm{kl}}_{\mathrm{n}}\right)\stackrel{\to }{\mathrm{c}}$ respectively then is a solution. ### Want to see more solutions like these? 