LaTeX Matter Types in Different Spatial Geometries

latentcorpse · Dec 16, 2009

hi i made the following table

 \begin{center} \begin{tabular}{l|l|l} \hline & \multicolumn{2}{c}{TYPE OF MATTER} \\ \cline{2-3} & ``Dust'' & ``Radiation'' \\ SPATIAL GEOMETRY & $P=0$ & $P=\frac{1}{3} \rho$ \\ \hline \\ \multirow{2}{*}{3-sphere, $K=1$} & $a=\frac{1}{2}C \left( 1 - \cos{\eta} \right) $ & $a=\sqrt{C'} \left( 1- \left( 1 - \frac{\tau}{\sqrt{C'}} \right)^2 \right)^{\frac{1}{2}}$ \\ & $\tau=\frac{1}{2}C \left( \eta - \sin{\eta} \right)$ & \\ \hline Flat, $k=0$ & $a= \left( \frac{9C}{4} \right)^{\frac{1}{3}} \tau^{\frac{2}{3}}$ & $a=\left( 4C' \right)^{\frac{1}{4}} \tau^{\frac{1}{2}}$ \\ \hline \multirow{2}{*}{3-hyperboloid, $k=-1$} & $a=\frac{1}{2} C \left( \cosh{\eta} - 1 \right)$ & $a=\sqrt{C'} \left( \left( 1 + \frac{\tau}{\sqrt{C'}} \right)^2 -1 \right)^{\frac{1}{2}}$ \\ & $\tau=\frac{1}{2}C \left( \sinh{\eta} - \eta \right)$ & \\ \hline \end{tabular} \end{center} 

with teh following code:

\begin{center}
\begin{tabular}{l|l|l}
\hline
& \multicolumn{2}{c}{TYPE OF MATTER} \\
\cline{2-3}
& ``Dust'' & ``Radiation'' \\
SPATIAL GEOMETRY & $P=0$ & $P=\frac{1}{3} \rho$ \\
\hline \\
\multirow{2}{*}{3-sphere, $K=1$} & $a=\frac{1}{2}C \left( 1 - \cos{\eta} \right) $ & $a=\sqrt{C'} \left( 1- \left( 1 - \frac{\tau}{\sqrt{C'}} \right)^2 \right)^{\frac{1}{2}}$ \\
& $\tau=\frac{1}{2}C \left( \eta - \sin{\eta} \right)$ & \\
\hline
Flat, $k=0$ & $a= \left( \frac{9C}{4} \right)^{\frac{1}{3}} \tau^{\frac{2}{3}}$ & $a=\left( 4C' \right)^{\frac{1}{4}} \tau^{\frac{1}{2}}$ \\
\hline
\multirow{2}{*}{3-hyperboloid, $k=-1$} & $a=\frac{1}{2} C \left( \cosh{\eta} - 1 \right)$ & $a=\sqrt{C'} \left( \left( 1 + \frac{\tau}{\sqrt{C'}} \right)^2 -1 \right)^{\frac{1}{2}}$ \\
& $\tau=\frac{1}{2}C \left( \sinh{\eta} - \eta \right)$ & \\
\hline
\end{tabular}
\end{center}

but i'd like any advice on how to make it look a bit more professional. for example, that line that's incomplete in the middle - how do i sort that?

thanks.

D H · Dec 16, 2009

Delete the '\\' after the \hline after the "SPATIAL GEOMETRY" line.

latentcorpse · Dec 16, 2009

cheers. i also get a line sticking out the bottom between the spatial geometry and dust columns. any advice on how to deal with that?

n.karthick · Aug 6, 2010

If you want "Spatial Geoemtry" to be in the middle vertically you can put this
\multirow{3}{*}{SPATIAL GEOMETRY}
in the first row itself.

Stalafin · Aug 22, 2010

Although n.karthick already necroposted, I guess, I can just add something as well:

Never EVER use the standard LaTeX tabular environment with vertical lines. It's absolutely horrible.

Use the booktabs package, and arrange your table layout by justifying your cells' contents.

JJackiw · Sep 13, 2010

The left side of A^* is A = \left[ \begin{smallmatrix} 9 &7\\ 1 &1 \end{smallmatrix} \right] and the right side of A^* is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right]. The idea is to reduce the left side of A^* from A to I in order to produce the right side of A^*_t from I to A^{-1}.

Sorry. Just testing. Apparently your testing thread is no longer active.

JJackiw · Sep 13, 2010

The left side of A^* is A = \left[ \begin{smallmatrix} 9 &7\\ 1 &1 \end{smallmatrix} \right] and the right side of A^* is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right]. The idea is to reduce the left side of A^* from A to I in order to produce the right side of A^*_t from I to A^{-1}.

Now the left side of A^*_t is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right] and the right side of A^*_t is A^{-1} = \left[ \frac{1}{2} \begin{smallmatrix} 1 &-7\\ -1 &9 \end{smallmatrix} \right]. In other words, as A is reduced from A to I, A^{-1} is simultaneously produced from I to A^{-1}.

JJackiw · Sep 13, 2010

The left side of A^* is A = \left[ \begin{smallmatrix} 9 &7\\ 1 &1 \end{smallmatrix} \right] and the right side of A^* is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right]. The idea is to reduce the left side of A^* from A to I in order to produce the right side of A^*_t from I to A^{-1}.

Now the left side of A^*_t is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right] and the right side of A^*_t is A^{-1} = \frac{1}{2} \left[ \begin{smallmatrix} 1 &-7\\ -1 &9 \end{smallmatrix} \right]. In other words, as A is reduced from A to I, A^{-1} is simultaneously produced from I to A^{-1}.

JJackiw · Sep 13, 2010

Your problem states that 325 people attended the theatre that day, grossing \$ 2675, at \$ 9 per ticket per adult and \$ 7 per ticket per child. This means that \$ 9 times the number of adults plus \$ 7 times the number of children produced \$ 2675, and that the total number of adults plus children was 325. This is your system of linear equations:

\begin{align*} 9x_1 + 7x_2 &= 2675\\ \phantom{9}x_1 + \phantom{7}x_2 &= 325 \end{align*}

You mentioned that you needed to use matrices to solve the problem. Then you will need to convert your system of linear equations to matrix form, as follows:

\begin{equation*} \begin{bmatrix} 9 & 7\\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} = \begin{bmatrix} 2675\\ 325 \end{bmatrix} \end{equation*}

This will produce a matrix equation of the form AX = B for coefficient matrix A = \left[ \begin{smallmatrix} 9 & 7\\ 1 & 1 \end{smallmatrix} \right], solution matrix X = \left[ \begin{smallmatrix} x_1\\ x_2 \end{smallmatrix} \right], constant matrix B = \left[ \begin{smallmatrix} 2675\\ 325 \end{smallmatrix} \right] and identity matrix I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right] such that:

\begin{align*} AX &= B\\ A^{-1}AX &= A^{-1}B\\ IX &= A^{-1}B\\ X &= A^{-1}B \end{align*}

Given X and B, you need to find A^{-1}, the inverse matrix of A, if it exists, to solve the equation. Using elementary row operations, r_i, on A^*, the augmented matrix of A, allows you to obtain A^{-1}:

\begin{align*} A^* &= [A|I] \xrightarrow{r_1} \cdots \xrightarrow{r_n} [I|A^{-1}] = A^*_t\\ &= \begin{bmatrix} 9 &7 &| &1 &0\\ 1 &1 &| &0 &1 \end{bmatrix} \xrightarrow{r_1} \cdots \xrightarrow{r_4} \begin{bmatrix} 1 &0 &| &\frac{1}{2} &-\frac{7}{2}\\ 0 &1 &| &-\frac{1}{2} &\frac{9}{2} \end{bmatrix} = A^*_t \end{align*} 
(Refer to the Appendix for a description of the elementary row operations which transform A^* to A^*_t.)

So that:

\begin{align*} A^{-1} &= \phantom{\dfrac{1}{2}} \begin{bmatrix} \frac{1}{2} &-\frac{7}{2}\\ -\frac{1}{2} &\frac{9}{2} \end{bmatrix}\\ &= \dfrac{1}{2} \begin{bmatrix} 1 &-7\\ -1 &9 \end{bmatrix} \end{align*}

Since X = A^{-1} B for X = \left[ \begin{smallmatrix} x_1\\ x_2 \end{smallmatrix} \right], A^{-1} = \frac{1}{2} \left[ \begin{smallmatrix} 1 &-7\\ -1 &9 \end{smallmatrix} \right] and B = \left[ \begin{smallmatrix} 2675\\ 325 \end{smallmatrix} \right], then:

\begin{align*} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} &= \dfrac{1}{2} \begin{bmatrix} 1 &-7\\ -1 &9 \end{bmatrix} \begin{bmatrix} 2675\\ 325 \end{bmatrix}\\ &= \dfrac{1}{2} \begin{bmatrix} 2675 - 7 \cdot 325\\ 9 \cdot 325 - 2675 \end{bmatrix}\\ &= \dfrac{1}{2} \begin{bmatrix} 2675-2275\\ 2925-2675 \end{bmatrix}\\ &= \phantom{\frac{1}{2}} \begin{bmatrix} 200\\ 125 \end{bmatrix} \end{align*}

Therefore, x_1=200 adults and x_2=125 children attended the theatre that day.

JJackiw · Sep 13, 2010

Appendix

The left side of A^* is A = \left[ \begin{smallmatrix} 9 &7\\ 1 &1 \end{smallmatrix} \right] and the right side of A^* is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right]. The idea is to reduce the left side of A^* from A to I in order to reproduce the right side of A^*_t from I to A^{-1}.

\begin{align*} A^* &= \; \, \begin{bmatrix} 9 &7 &|&1 &0\\ 1 &1 &|&0 &1 \end{bmatrix}\\ &\xrightarrow{r_1} \begin{bmatrix} 1 &1 &| &0 &1\\ 9 &7 &| &1 &0 \end{bmatrix}\\ &\xrightarrow{r_2} \begin{bmatrix} 1 &1 &| &0 &1\\ 0 &-2 &| &1 &-9 \end{bmatrix}\\ &\xrightarrow{r_3} \begin{bmatrix} 1 &1 &| &0 &1\\ 0 &1 &| &-\frac{1}{2} &\frac{9}{2} \end{bmatrix}\\ &\xrightarrow{r_4} \begin{bmatrix} 1 &0 &| &\frac{1}{2} &-\frac{7}{2}\\ 0 &1 &| &-\frac{1}{2} &\frac{9}{2} \end{bmatrix}\\ &\; \, = A^*_t \end{align*}

Now the left side of A^*_t is I = \left[ \begin{smallmatrix} 1 &0\\ 0 &1 \end{smallmatrix} \right] and the right side of A^*_t is A^{-1} = \left[ \begin{smallmatrix} 1/2 &-7/2\\ -1/2 &9/2 \end{smallmatrix} \right]. In other words, as A is reduced from A to I, A^{-1} is simultaneously reproduced from I to A^{-1}.

LaTeX Matter Types in Different Spatial Geometries

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