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\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

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\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.

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