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\input{../preamble.tex}



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\begin{document}

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\title{\normalsize \textsc{}
		\\ [2.0cm]
		\HRule{1.5pt} \\
		\LARGE \textbf{\uppercase{MA1522}
		\HRule{2.0pt} \\ [0.6cm] \LARGE{Assignment 1} \vspace*{10\baselineskip}}
		}
\date{}
\author{\textbf{Yadunand Prem}\\
A0253252M}

\maketitle
\newpage

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\section{Question 1}
\hr
\begin{align*}
  \begin{amatrix}{3}
    4 & 2 & 4 & 0 \\
    5 & 4 & 0 & 1 \\
    4 & 1 & 2 & 5 \\
  \end{amatrix}
  \xrightarrow{\frac{1}{4}R_1}
  & \begin{amatrix}{3}
    1 & \frac{1}{2} & 1 & 0 \\
    5 & 4 & 0 & 1 \\
    4 & 1 & 2 & 5 \\
  \end{amatrix}
  \xrightarrow[R_3 + (-4)R_1]{R_2 + (-5)R_1}
  \begin{amatrix}{3}
    1 & \frac{1}{2} & 1  & 0 \\
    0 & \frac{3}{2} & -5 & 1 \\
    0 & -1          & -2 & 5 \\
  \end{amatrix} \\
  \xrightarrow{\frac{2}{3}R_2}
  & \begin{amatrix}{3}
    1 & \frac{1}{2} & 1  & 0 \\
    0 & 1 & \frac{-10}{3} & \frac{2}{3} \\
    0 & -1          & -2 & 5 \\
  \end{amatrix}
  \xrightarrow{R_3 + R_2}
  \begin{amatrix}{3}
    1 & \frac{1}{2} & 1  & 0 \\
    0 & 1 & \frac{-10}{3} & \frac{2}{3} \\
    0 & 0 & \frac{-16}{3} & \frac{17}{3} \\
  \end{amatrix} \\
  \xrightarrow{\frac{-3}{16}R_3}
  & \begin{amatrix}{3}
    1 & \frac{1}{2} & 1  & 0 \\
    0 & 1 & \frac{-10}{3} & \frac{2}{3} \\
    0 & 0 & 1 & \frac{-17}{16} \\
  \end{amatrix}
  \xrightarrow[R_1 + (-1)R_3]{R_2 + \frac{10}{3}R_3}
  \begin{amatrix}{3}
    1 & \frac{1}{2} & 0  & \frac{17}{16} \\
    0 & 1 & 0 & \frac{-23}{8} \\
    0 & 0 & 1 & \frac{-17}{16} \\
  \end{amatrix} \\
  \xrightarrow{R_1 + \frac{-1}{2}R_2}
  & \begin{amatrix}{3}
    1 & 0 & 0 & \frac{5}{2} \\
    0 & 1 & 0 & \frac{-23}{8} \\
    0 & 0 & 1 & \frac{-17}{16} \\
  \end{amatrix}
\end{align*}

\subsection{1i No of pivot columns}
There are 3 pivot columns

\subsection{1ii Arbitrary Params needed}
0 arbitrary params needed

\subsection{1iii How many solutions are there?}
1 solution for the system

\subsection{1iv Solution for system}
$x_1 = \frac{5}{2}, x_2 = -\frac{23}{8}, x_3 = -\frac{17}{16}$
\newpage

\section{Question 2}
\begin{align*}
  \begin{amatrix}{3}
    a & 1 & 1 & a^3 \\
    1 & a & 1 & 1 \\
    1 & 1 & a & a
  \end{amatrix}
  \xrightarrow{R_1 \leftrightarrow R_3}
  & \begin{amatrix}{3}
    1 & 1 & a & a \\
    1 & a & 1 & 1 \\
    a & 1 & 1 & a^3 \\
  \end{amatrix} \\
  \xrightarrow[R_3 + (-a)R_1]{R_2 + (-1)R_1}
  & \begin{amatrix}{3}
    1 & 1 & a & a \\
    0 & a-1 & 1-a & 1-a \\
    0 & 1-a & 1-a^2 & a^3-a^2 \\
  \end{amatrix} \\
  \xrightarrow{R_3 + (1)R_2}
  & \begin{amatrix}{3}
    1 & 1 & a & a \\
    0 & a-1 & 1-a & 1-a \\
    0 & 0 & 2-a^2-a & a^3-a^2-a+1 \\
  \end{amatrix} \\
  = 
  & \begin{amatrix}{3}
    1 & 1 & a & a \\
    0 & a-1 & 1-a & 1-a \\
    0 & 0 & (a+2)(1-a) & (a+1)(a-1)(a-1) \\
  \end{amatrix} \\
\end{align*}

\subsection{2i, system has no solution}
The system will have no solution when the last column is the pivot column. This can happen in $R_2$ if $a-1 = 1-a = 0$ and $1-a \neq 0$

$a = 1$ will satisfy the left equation, but it contradicts with the right equation. Thus, $R_2$ is not a possible candidate for no solution

For $R_3$, $(a+2)(1-a) = 0$ and $(a+1)(a-1)(a-1) \neq 0$

Left equation: 
\begin{align*}
(a+2)(1-a) = 0 \\
a = 1 \text{ or } a = -2
\end{align*}

Right Equation: 
\begin{align*}
  (a+1)(a-1)(a-1) \neq 0 \\
  a \neq 1 \text{ or }a \neq -1 
\end{align*}

Therefore, there is no solution to the equation if $a = -2$

\subsection{2ii, system has unique solution}
The system has a unique solution when the last column is non pivot and all other columns are pivot columns. 

$C_1$ is pivot column, $C_2$ is pivot if $a-1 \neq 0$, $C_3$ is pivot if $(a+2)(1-a) \neq 0$

\begin{align*}
  a - 1 \neq 0\\ 
  a \neq 1 \\
  (a+2)(1-a) \neq 0 \\
  a \neq -2 \text{ and } a \neq 1
\end{align*}

Therefore, the system has a unique solutions when $a \neq 1$ and $a \neq -2$

\subsection{2iii System has infinitely many solutions}
The system will have infinitely many solutions when the last column is non-pivot and some other columns are non-pivot columns. 

if $C_2$ is non pivot, then $C_2R_2$ must be 0. $a-1 = 0, a = 1$

If $C_3$ is non-pivot, then $C_3R_3$ must be zero, but that will cause the last row to be a pivot. 

Therefore, the system has infinitely many solutions when $a = 1$

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