# Matter Types in Different Spatial Geometries

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• latentcorpse
In summary, the system of linear equations represents the problem of finding the number of adults and children who attended the theatre and the amount of money earned based on ticket prices. It can be solved by converting it into a matrix equation and finding the inverse of the coefficient matrix. Using elementary row operations, the inverse matrix can be obtained and used to solve for the values of x1 and x2. The solution shows that 200 adults and 125 children attended the theatre, earning a total of $2675. latentcorpse 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.

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

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?

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

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.

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 $A^{-1}$.

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 $A^{-1}$.

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 $A^{-1}$.

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.

Last edited:
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 $A^{-1}$.

## 1. What is matter and why is it important to study different types in different spatial geometries?

Matter is anything that has mass and takes up space. It is important to study different types of matter in different spatial geometries because it allows us to understand the physical properties and behavior of matter in various environments. This knowledge is crucial for many fields of science, such as materials science, chemistry, and astrophysics.

## 2. What are the three states of matter?

The three states of matter are solid, liquid, and gas. These states are determined by the arrangement and movement of particles in a substance. Solids have tightly packed particles that vibrate in place, liquids have particles that can move around and slide past each other, and gases have particles that move freely and are widely spaced.

## 3. How does matter behave in different spatial geometries, such as in a vacuum or in microgravity?

In different spatial geometries, matter can behave differently due to factors such as gravity and pressure. In a vacuum, where there is no air or other matter, particles may behave differently than in a normal environment. In microgravity, where there is very little gravity, particles may have different interactions and properties compared to on Earth.

## 4. What are the different types of matter in terms of their composition?

Matter can be classified into two main types: pure substances and mixtures. Pure substances are made up of only one type of atom or molecule, while mixtures are made up of two or more different substances. Mixtures can be further categorized as homogenous (uniform composition) or heterogenous (non-uniform composition).

## 5. How does the shape of an object affect the behavior of its matter?

The shape of an object can affect the behavior of its matter in various ways. For example, the shape of a container can determine how a liquid or gas is able to move and fill the space inside. The shape of a solid object can also impact its strength, density, and other physical properties. In different spatial geometries, such as in microgravity, the shape of an object can also affect how it moves and interacts with other objects due to the absence of normal gravitational forces.

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