coursetitle={Physics 102},
classtitle={Linear algebra introduction},
subheading={Solving systems of linear equations automatically},
- loose]{../../tex/problempack}
+ loose]{problempack}
+\usepackage{wtk_cmmds}
%-----------------------end preamble-------------------------
\section{Writing matrix equations}
-All battery-resistor-current problems produce systems of \emph{linear equations}.
-Linear equations are just a equations where the unknowns (usually the currents $I_i$) are never multiplied together.
-For example from Recitation 6, Problem 35, we have
+All battery-resistor-current problems produce systems of \emph{linear
+ equations}. Linear equations are just a equations where the
+unknowns (usually the currents $I_i$) are never multiplied together.
+For example, from Serway and Jewett's 4$^\text{th}$ Edition, Problem
+21.35, we have
\begin{align}
0 &= I_1 - I_2 - I_3 \\
0 &= \epsilon_2 - I_2 R_2 - I_1 R_1 \\
0 &= \epsilon_3 - I_3 R_3 - I_1 R_1 \;,
\end{align}
-from Kirchhoff's laws.
-We're given everything except the currents, and no terms have two currents multiplied by each-other in them.
-We can rewrite these equations and adjust the spacing a bit to get
+from Kirchhoff's laws. We're given everything except the currents,
+and no terms have two currents multiplied by each-other in them. We
+can rewrite these equations and adjust the spacing a bit to get
\begin{align}
0 &= -I_1 \;\;+\;\; I_2 \;\;+\;\; I_3 \label{eq.K1} \\
\epsilon_2 &= R_1 I_1 + R_2 I_2 + 0\cdot I_3 \\
\end{align}
We can see that each line has the same format
\begin{equation}
- b = a_1 I_1 + a_2 I_2 + a_3 I_3 = \sum_{j=1}^3 a_j I_j \;,
+ b = a_1 I_1 + a_2 I_2 + a_3 I_3 = \sum_{i=1}^3 a_i I_i \;,
\end{equation}
with constant $a_i$s and $b$s.
I_1 \\
I_2 \\
I_3
- \end{pmatrix} \label{eq.KM}
+ \end{pmatrix} \;. \label{eq.KM}
\end{equation}
-Comparing this to Eqns.~\ref{eq.K1}--\ref{eq.K3}, we can see that all we've done is erased a few equals signs in the middle, drawn some big parenthesis, and dragged the unknown currents off by themselves, giving them their own parenthesis.
-We've also rotated order we wrote the currents in
+Comparing this to Eqns.~\ref{eq.K1}--\ref{eq.K3}, we can see that all
+we've done is erased a few equals signs in the middle, drawn some big
+parenthesis, and dragged the unknown currents off by themselves,
+giving them their own parenthesis. We've also rotated order we wrote
+the currents in
\begin{equation}
\clubsuit I_1 + \diamondsuit I_2 + \heartsuit I_3
\longrightarrow
I_3
\end{pmatrix} \;.
\end{equation}
-Things look fairly different at first, so go back and compare Eqn.~\ref{eq.KM} with Eqns.~\ref{eq.K1}--\ref{eq.K3} until you get comfortable with the changes.
+Things look fairly different at first, so go back and compare
+Eqn.~\ref{eq.KM} with Eqns.~\ref{eq.K1}--\ref{eq.K3} until you get
+comfortable with the changes.
\section{Solving matrix equations}
-All the things we could do when we had regular equations, we can still do with the equations written in matrix form. We'll go through and solve these side by side so you can see a solution in action and become more familiar with the matrix notation.
-The only difference in solving strategy is that we keep all the symbols on the same side they started on, and add or subtract equations instead of plugging in.
+All the things we could do when we had regular equations, we can still
+do with the equations written in matrix form. We'll go through and
+solve these side by side so you can see a solution in action and
+become more familiar with the matrix notation. The only difference in
+solving strategy is that we keep all the symbols on the same side they
+started on, and add or subtract equations instead of plugging in.
\begin{align}
\begin{matrix}
0 &= &-I_1 &+ &I_2 &+ &I_3 \\
I_3
\end{pmatrix}
\end{align}
-Now we can subtract (just like plugging in) the second equation to the first, to get rid of the $I_2$ in the first equation in terms of $I_1$.
+Now we can subtract (just like plugging in) the second equation to the
+first, to get rid of the $I_2$ in the first equation in terms of
+$I_1$.
\begin{align}
\begin{matrix}
\frac{-\epsilon_2}{R_2} &= &-\p({\frac{R_1}{R_2}+1})I_1 &+ &0 &+ &I_3 \\
I_3
\end{pmatrix}
\end{align}
-\ldots and subtracting the last equation to the first, to get rid of the $I_3$ in the first equation in terms of $I_1$.
+\ldots and subtracting the last equation to the first, to get rid of
+the $I_3$ in the first equation in terms of $I_1$.
\begin{align}
\begin{matrix}
- \frac{-\epsilon_3}{R_3}-\frac{\epsilon_2}{R_2} &= &-\p({\frac{R_1}{R_3}+\frac{R_1}{R_2}+1})I_1 &+ &0 &+ &0 \\
+ \frac{-\epsilon_3}{R_3}-\frac{\epsilon_2}{R_2}
+ &= &-\p({\frac{R_1}{R_3}+\frac{R_1}{R_2}+1})I_1 &+ &0 &+ &0 \\
\frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\
\frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3
\end{matrix}
I_3
\end{pmatrix}
\end{align}
-Now we have an equation with just $I_1$ and the $R_i$ and $\epsilon_i$ that were given.
-Dividing through by the junk in front of $I_1$ we have
+Now we have an equation with just $I_1$ and the $R_i$ and $\epsilon_i$
+that were given. Dividing through by the junk in front of $I_1$ we
+have
\begin{align}
\begin{matrix}
- \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\
+ \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}
+ {\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\
\frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\
\frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3
\end{matrix}
&&
\begin{pmatrix}
- \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\
+ \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}
+ {\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\
\frac{\epsilon_2}{R_2} \\
\frac{\epsilon_3}{R_3}
\end{pmatrix}
I_3
\end{pmatrix}
\end{align}
-A solution! We know have an explicit expression for $I_1$.
-Going back and subtracting $R_1/R_2$ time the first equation from the middle lets us solve for $I_2$.
+A solution! We know have an explicit expression for $I_1$. Going
+back and subtracting $R_1/R_2$ time the first equation from the middle
+lets us solve for $I_2$.
\begin{align}
\begin{matrix}
- \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\
- \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} &= &0 &+ &I_2 &+ &0 \\
+ \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}
+ {\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\
+ \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}
+ +\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}}
+ &= &0 &+ &I_2 &+ &0 \\
\frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3
\end{matrix}
&&
\begin{pmatrix}
- \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\
- \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} \\
+ \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}
+ {\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\
+ \frac{\epsilon_2}{R_2}
+ - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}
+ {\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} \\
\frac{\epsilon_3}{R_3}
\end{pmatrix}
=
I_3
\end{pmatrix}
\end{align}
-And finally subtracting $R_1/R_3$ time the first equation from the last lets us solve for $I_3$.
+Finally subtracting $R_1/R_3$ time the first equation from the last
+lets us solve for $I_3$.
\begin{align}
\begin{matrix}
\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\
\end{pmatrix}
\end{align}
-Yuck! Why would anyone want to solve equations like that?
-You have to do all the same work in the matrix version, but you're rewriting everything at each step!
-What gives?
-The power of linear algebra is that, while it is long and tedious for us (see above), it is very easy for a computer (see below).
-Understanding why means taking a step back and getting an abstract view of the solution we just worked through.
+Yuck! Why would anyone want to solve equations like that? You have
+to do all the same work in the matrix version, but you're rewriting
+everything at each step! What gives? The power of linear algebra is
+that, while it is long and tedious for us (see above), it is very easy
+for a computer (see below). Understanding why means taking a step
+back and getting an abstract view of the solution we just worked
+through.
\section{Solving matrix equations with a computer}
I
\end{bmatrix}
\end{equation}
-Wait a second, the $\epsilon$s were voltages, so that looks a lot like plain old $V=IR$!
-We know how to get the current then, it's just
+Wait a second, the $\epsilon$s were voltages, so that looks a lot like
+plain old $V=IR$! We know how to get the current then, it's just
\begin{equation}
R^{-1} V = I \;.
\end{equation}
-Maybe we can think about our matrix equation like this, we just need to figure out what $[R]^{-1}$ means.
-What is $R^{-1}$ anyway?
-It's the \emph{inverse} of $R$; the thing that, when multiplied by $R$, gives one.
-$[R]^{-1}$ is also just the inverse of $[R]$, so $[R]^{-1}\cdot[R] = 1$.
+Maybe we can think about our matrix equation like this, we just need
+to figure out what $[R]^{-1}$ means. What is $R^{-1}$ anyway? It's
+the \emph{inverse} of $R$; the thing that, when multiplied by $R$,
+gives one. $[R]^{-1}$ is also just the inverse of $[R]$, so
+$[R]^{-1}\cdot[R] = 1$.
-Inverting a matrix by hand is basically what we were doing in our solution above,
-but if we can get our computer (or calculator) to find $[R]^{-1}$ for us, we don't have to do any of the messy algebra.
-We can just get our solution via
+Inverting a matrix by hand is basically what we were doing in our
+solution above, but if we can get our computer (or calculator) to find
+$[R]^{-1}$ for us, we don't have to do any of the messy algebra. We
+can just get our solution via
\begin{equation}
\begin{bmatrix}
R
\end{bmatrix}
\end{equation}
-On the TI-83+, that's pretty much all there is to it.
-You can enter your $[R]$ and $[\epsilon]$ matrices in the {\tt [2nd] MATRX $\rightarrow$ EDIT} menu (calling them $[A]$ and $[B]$.
-Then just type out
+On the TI-83+, that's pretty much all there is to it. You can enter
+your $[R]$ and $[\epsilon]$ matrices in the {\tt [2nd] MATRX
+ $\rightarrow$ EDIT} menu (calling them $[A]$ and $[B]$. Then just
+type out
\begin{equation}
- [A]^{-1}*B
+ [A]^{-1}*[B]
\end{equation}
-using the {\tt [2nd] MATRX $\rightarrow$ NAMES} menu to generate the $[A]$ and $[B]$ symbols.
+using the {\tt [2nd] MATRX $\rightarrow$ NAMES} menu to generate the
+$[A]$ and $[B]$ symbols.
-On the TI-89, you can enter the matrices straight from the command line, using comas to separate the columns and semi-colons to separate the rows. Using the numbers from Problem 35 that gives
+On the TI-89, you can enter the matrices straight from the command
+line, using comas to separate the columns and semi-colons to separate
+the rows. Using the numbers from Serway and Jewett's 4$^\text{th}$
+Edition, Problem 21.35, that gives
\begin{align}
&[-1,1,1;8,6,0;8,0,4] \rightarrow A \\
&[0;4;12] \rightarrow I \\
\end{align}
(I don't have a TI-89, so if this is wrong, let me know\ldots).
-
\end{document}
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