feat: add 1522 assignment

ma1522/assignment/assignment1.tex

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+\documentclass[a4paper]{article}
+
+\input{../preamble.tex}
+
+
+
+% ------------------------------------------------------------------------------
+
+\begin{document}
+
+% ------------------------------------------------------------------------------
+% Cover Page and ToC
+% ------------------------------------------------------------------------------
+
+\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
+
+% ------------------------------------------------------------------------------
+
+\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 $a = 1$
+
+% ------------------------------------------------------------------------------
+
+\end{document}