• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q21E

Expert-verified Found in: Page 439 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Ngozi opens a bank account with an initial balance of 1,000 Nigerian naira. Let b(t) be the balance in the account at time t; we are told that b(0) = 1,000. The bank is paying interest at a continuous rate of 5% per year. Ngozi makes deposits into the account at a continuous rate of s(t) (measured in naira per year). We are told that s(0) = 1,000 and that s(t) is increasing at a continuous rate of 7% per year. (Ngozi can save more as her income goes up over time.) (a) set up a linear system of the form $\mathbf{|}\begin{array}{c}\frac{\mathbf{d}\mathbf{b}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{?}\mathbf{b}\mathbf{+}\mathbf{?}\mathbf{s}\\ \frac{\mathbf{d}\mathbf{s}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{?}\mathbf{b}\mathbf{+}\mathbf{?}\mathbf{s}\end{array}\mathbf{|}$(b) Find ${\mathbit{b}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$and $\mathbit{s}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$.

(a) The solution is $|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}|$

(b) The solution$b\left(t\right)=-49000{e}^{0.05t}+50000{c}_{2}{e}^{0.07t},\text{\hspace{0.17em}\hspace{0.17em}}s\left(t\right)=1000{e}^{0.07t}$

See the step by step solution

## Step1: Explanation of the solution

Let$b\left(t\right)$ is the balance in the account at time $t$ and $b\left(0\right)=1000$.

The bank is paying interest at a continuous rate of $5%$ per year.

Ngozi makes deposits in to the account at a continuous rate of$s\left(t\right)$ .

And $s\left(t\right)$is increasing at a continuous rate of $7%$ per year also that $s\left(0\right)=1000$.

The setup of a linear system as follows.

$|\begin{array}{c}\frac{db}{dt}=?b+?s\\ \frac{ds}{dt}=?b+?s\end{array}|$

Since, the bank is already paying interest at a continuous rate of $5%$ per year and Ngozi makes deposits into the account at a continuous rate of$s\left(t\right)$ per year.

The first equation can written as follows.

$|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=?b+?s\end{array}|$

Since, $s\left(t\right)$ is increasing at a continuous rate of$7%$ per year.

Therefore, the equation can be written as follows.

$|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}|$

Hence, the setup of the linear system is $|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}|$

## Step 2: Explanation of the solution

Consider the linear system of equation as follows.

$|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}|$

Rewrite the system in the form of $\frac{d\stackrel{\to }{x}}{dt}=A\stackrel{\to }{x}$ as follows.

$\left[\begin{array}{c}\frac{db}{dt}\\ \frac{ds}{dt}\end{array}\right]=\left[\begin{array}{cc}0.05& 1\\ 0& 0.07\end{array}\right]\left[\begin{array}{c}b\\ s\end{array}\right]$

Since the coefficient matrix $A=\left[\begin{array}{cc}0.05& 1\\ 0& 0.07\end{array}\right]$ is a diagonal matrix, the eigenvalues are the diagonal entries.

The eigenvalues are as follows.

$\begin{array}{l}{\lambda }_{1}=0.05\\ {\lambda }_{2}=0.07\end{array}$

Now, find the corresponding eigenvectors for eigenvalue${\lambda }_{1}=0.05$ as follows.

$\begin{array}{c}\left(A-\lambda I\right)=0\\ \left[\begin{array}{cc}0.05-\left(0.05\right)& 1\\ 0& 0.07-\left(0.05\right)\end{array}\right]\left[\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 00\end{array}\right]\\ \left[\begin{array}{cc}0& 1\\ 0& 0.02\end{array}\right]\left[\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 00\end{array}\right]\\ \left[\begin{array}{c}0{u}_{1}+1{u}_{2}\\ 0{u}_{1}+0.02{u}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

The corresponding equations are as follows.

$\begin{array}{c}0{u}_{1}+1{u}_{2}=0\\ 0{u}_{1}+0.2{u}_{2}=0\end{array}$

Solving the equations as follows.

$\begin{array}{c}0{u}_{1}+1{u}_{2}=0\\ {u}_{2}=-0{u}_{1}\\ {u}_{2}=0\end{array}$

Therefore, the eigenvectors as follows.

$\begin{array}{c}\stackrel{\to }{u}=\left[\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right]\\ =\left[\begin{array}{c}{u}_{1}\\ 0\end{array}\right]\\ \stackrel{\to }{u}={u}_{1}\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Similarly, find the corresponding eigenvectors for eigenvalue ${\lambda }_{1}=0.07$as follows.

$\begin{array}{c}\left(A-\lambda I\right)=0\\ \left[\begin{array}{cc}0.05-\left(0.07\right)& 1\\ 0& 0.07-\left(0.07\right)\end{array}\right]\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}0.02& 1\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{c}0.02{v}_{1}+1{v}_{2}\\ 0{v}_{1}+0{v}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

The corresponding equations are as follows.

$\begin{array}{c}0.02{v}_{1}+1{v}_{2}=0\\ 0{v}_{1}+0{v}_{2}=0\end{array}$

Solving the equations as follows.

$\begin{array}{c}0.02{v}_{1}+1{v}_{2}=0\\ 0.02{v}_{1}=-{v}_{2}\\ {v}_{1}=\frac{-1{v}_{2}}{0.02}\\ {v}_{1}=50{v}_{2}\end{array}$

Therefore, the eigenvectors as follows.

$\begin{array}{c}\stackrel{\to }{v}=\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]\\ =\left[\begin{array}{c}50{v}_{2}\\ {v}_{2}\end{array}\right]\\ \stackrel{\to }{v}={v}_{2}\left[\begin{array}{c}50\\ 1\end{array}\right]\end{array}$

Thus, the eigenvectors are as follows.

$\left[\begin{array}{c}1\\ 0\end{array}\right]\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{c}50\\ 1\end{array}\right]$

The general solution to the system can be written as follows.

$\begin{array}{c}\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]={c}_{1}{e}^{{\lambda }_{1}t}\stackrel{\to }{u}+{c}_{2}{e}^{{\lambda }_{2}t}\stackrel{\to }{v}\\ ={c}_{1}{e}^{0.05t}\left[\begin{array}{c}1\\ 0\end{array}\right]+{c}_{2}{e}^{0.07t}\left[\begin{array}{c}50\\ 1\end{array}\right]\\ =\left[\begin{array}{c}{c}_{1}{e}^{0.05t}\\ 0\end{array}\right]+\left[\begin{array}{c}50{c}_{2}{e}^{0.07t}\\ {c}_{2}{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_{1}{e}^{0.05t}+50{c}_{2}{e}^{0.07t}\\ {c}_{2}{e}^{0.07t}\end{array}\right]\end{array}$

Now substitute the value 0 for in$\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_{1}{e}^{0.05t}+50{c}_{2}{e}^{0.07t}\\ {c}_{2}{e}^{0.07t}\end{array}\right]$ as follows.

$\begin{array}{l}\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_{1}{e}^{0.05t}+50{c}_{2}{e}^{0.07t}\\ {c}_{2}{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(0\right)\\ s\left(0\right)\end{array}\right]=\left[\begin{array}{c}{c}_{1}{e}^{0.05\left(0\right)}+50{c}_{2}{e}^{0.07\left(0\right)}\\ {c}_{2}{e}^{0.07\left(0\right)}\end{array}\right]\\ \left[\begin{array}{c}1000\\ 1000\end{array}\right]=\left[\begin{array}{c}{c}_{1}+50{c}_{2}\\ {c}_{2}\end{array}\right]\end{array}$

Now, equating the corresponding entries of the matrix on sides of the equation as follows.

$\begin{array}{c}{c}_{1}+50{c}_{2}=1000\\ {c}_{2}=1000\end{array}$

Substitute the value $1000$ for${c}_{2}$ in ${c}_{1}+50{c}_{2}=1000$as follows.

$\begin{array}{c}{c}_{1}+50{c}_{2}=1000\\ {c}_{1}+50\left(1000\right)=1000\\ {c}_{1}+50000=1000\\ {c}_{1}=1000-50000\end{array}$

Simplify further as follows.

${c}_{1}=-49,000$

Substitute the value$-49,000$ for${c}_{1}$ and$1000$ for${c}_{2}$ in$\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_{1}{e}^{0.05t}+50{c}_{2}{e}^{0.07t}\\ {c}_{2}{e}^{0.07t}\end{array}\right]$ as follows.

$\begin{array}{l}\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_{1}{e}^{0.05t}+50{c}_{2}{e}^{0.07t}\\ {c}_{2}{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}-49000{e}^{0.05t}+50\left(1000\right){c}_{2}{e}^{0.07t}\\ 1000{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}-49000{e}^{0.05t}+50000{c}_{2}{e}^{0.07t}\\ 1000{e}^{0.07t}\end{array}\right]\end{array}$

Equating the corresponding entries on both sides of the equation as follows.

$\begin{array}{l}b\left(t\right)=-49000{e}^{0.05t}+50000{c}_{2}{e}^{0.07t}\\ s\left(t\right)=1000{e}^{0.07t}\end{array}$

Thus, the equations are $b\left(t\right)=-49000{e}^{0.05t}+50000{c}_{2}{e}^{0.07t},\text{\hspace{0.17em}\hspace{0.17em}}s\left(t\right)=1000{e}^{0.07t}$. ### Want to see more solutions like these? 