feat: 1522 assignment 2

ma1522/assignment/assignment2.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 2} \vspace*{10\baselineskip}}
+		}
+\date{}
+\author{\textbf{Yadunand Prem}\\
+A0253252M}
+
+\maketitle
+\newpage
+
+% ------------------------------------------------------------------------------
+
+\section{Question 1}
+\hr
+\begin{align*}
+  \overrightarrow{A} = \begin{pmatrix}
+    2 & -4 & 4 & -2 \\
+    6 & a-12 & 7 & a-6 \\
+    -1 & 2-b & 8 & -b+1 \\
+  \end{pmatrix}
+  \xrightarrow[R_3 + \frac{1}{2}R_1]{R_2 + (-3)R_1}
+  & \begin{pmatrix}
+    2 & -4 & 4 & -2 \\
+    0 & a & -5 & a \\
+    0 & -b & 10 & -b \\
+  \end{pmatrix} \\
+  \xrightarrow{R_3 + \frac{b}{a}R_2}
+  & \begin{pmatrix}
+    2 & -4 & 4 & -2 \\
+    0 & a & -5 & a \\
+    0 & 0 & 10-\frac{5b}{a} & 0 \\
+  \end{pmatrix} = \overrightarrow{U}
+\end{align*}
+
+\begin{align*}
+  \overrightarrow{I} \xrightarrow{R_3 + \frac{-b}{a}R_2} \xrightarrow[R_3 + \frac{-1}{2}R_1]{R_2 + (3)R_1} \overrightarrow{L}
+\end{align*}
+
+\begin{align*}
+  \begin{pmatrix}
+    1 & 0 & 0 \\
+    0 & 1 & 0 \\
+    0 & 0 & 1
+  \end{pmatrix}
+  \xrightarrow{R_3 + (-\frac{b}{a})R_2}
+  \begin{pmatrix}
+    1 & 0 & 0 \\
+    0 & 1 & 0 \\
+    0 & -\frac{b}{a} & 1
+  \end{pmatrix}
+  \xrightarrow[R_3 + \frac{-1}{2}R_1]{R_2 + (3)R_1}
+  \begin{pmatrix}
+    1 & 0 & 0 \\
+    3 & 1 & 0 \\
+    -\frac{1}{2} & -\frac{b}{a} & 1
+  \end{pmatrix} = \overrightarrow{L}
+\end{align*}
+If $a = 0$, then in the 1st step of the factorization above, it will cause the array to pivot. Thus, to prevent pivoting, $b = 0$ must be the case also.
+$a \neq 0$ then $b \neq 0$ also, as $\frac{b}{a}$ will not be defined
+
+\newpage
+\section{Question 2}
+\hr
+For A to be invertible, $det(A) \neq 0$. 
+
+$A = \begin{vmatrix}
+  1 & 1 & 1 & 1  \\
+  1 & 2 &-1 & c  \\
+  1 & 4 & 1 & c^2 \\
+  1 & 8 &-1 & c^3 \\
+\end{vmatrix}
+$
+
+\begin{align*}
+  det(A) 
+  & = (1)\begin{vmatrix} 
+    2 &-1 & c  \\
+    4 & 1 & c^2 \\
+    8 &-1 & c^3 \\
+  \end{vmatrix} + 
+  (-1)\begin{vmatrix}
+  1  &-1 & c  \\
+  1  & 1 & c^2 \\
+  1  &-1 & c^3 \\
+\end{vmatrix} + 
+(1)\begin{vmatrix}
+  1 & 2 & c  \\
+  1 & 4 & c^2 \\
+  1 & 8 & c^3 \\
+\end{vmatrix} + 
+(-1)\begin{vmatrix}
+  1 & 2 &-1 \\
+  1 & 4 & 1 \\
+  1 & 8 &-1 \\
+\end{vmatrix} \\
+  & = (1)\left((2)\begin{vmatrix} 
+     1 & c^2 \\
+    -1 & c^3 \\
+  \end{vmatrix} + 
+  (1)\begin{vmatrix} 
+    4 & c^2 \\
+    8 & c^3 \\
+  \end{vmatrix} + 
+  (c)\begin{vmatrix} 
+    4 & 1  \\
+    8 &-1  \\
+  \end{vmatrix}\right)\\
+  & + (-1)\left((1)\begin{vmatrix} 
+   1 & c^2 \\
+  -1 & c^3 \\
+  \end{vmatrix} + 
+  (1)\begin{vmatrix} 
+  1  & c^2 \\
+  1  & c^3 \\
+  \end{vmatrix} + 
+  (c)\begin{vmatrix} 
+  1  & 1 \\
+  1  &-1 \\
+  \end{vmatrix}\right)\\
+  & + (1)\left((1)\begin{vmatrix} 
+  4 & c^2 \\
+  8 & c^3 \\
+  \end{vmatrix} + 
+  (-2)\begin{vmatrix} 
+  1 & c^2 \\
+  1 & c^3 \\
+  \end{vmatrix} + 
+  (c)\begin{vmatrix} 
+  1 & 4 \\
+  1 & 8 \\
+  \end{vmatrix}\right)\\
+  & + (-1)\left((1)\begin{vmatrix} 
+  4 & 1 \\
+  8 &-1 \\
+  \end{vmatrix} + 
+  (-2)\begin{vmatrix} 
+  1 & 1 \\
+  1 &-1 \\
+  \end{vmatrix} + 
+  (-1)\begin{vmatrix} 
+  1 & 4 \\
+  1 & 8 \\
+  \end{vmatrix}\right)\\
+  & = (2)(c^3+c^2) + (4c^3-8c^2) + (c)(-4-8)\\
+  & + (-1)((c^3+c^2) + (c^3-c^2) + (c)(-1 - 1))\\
+  & + (4c^3-8c^2) + (-2)(c^3-c^2) + (c)(8-4)\\
+  & + (-1)((-4-8) + (-2)(-1-1) + (-1)(8-4)\\
+  & = 6c^3 - 12c^2 - 6c + 12 \\
+  & = 6(c - 2)(c - 1)(c + 1)
+\end{align*}
+
+Therefore, $det(A) = (c-2)(c-1)(c+1) \neq 0$,  \\
+$c \neq 2, c \neq 1, c \neq -1$
+
+Actually, this question can be solved by observation. The determinant will be 0 if 2 columns are a multiple of one another. By observation, we can see that if $c = 1$, then the 4th and 1st column will be the same. If $c = 2$, then 2,4 and $c=-1$, then 3,4 will be the same.
+
+% ------------------------------------------------------------------------------
+
+\end{document}