614 lines
19 KiB
TeX
614 lines
19 KiB
TeX
\subsection{Linear Algebra}
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\begin{itemize}
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\item \textbf{Linear} The study of items/planes and objects which are flat
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\item \textbf{Algebra} Objects are not as simple as numbers
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\end{itemize}
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\subsection{Linear Systems \& Their Solutions}
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Points on a straight line are all the points $(x, y)$ on the $xy$ plane satisfying the linear eqn: $ax + by = c$, where $a, b > 0$
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\subsubsection{Linear Equation}
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Linear eqn in $n$ variables (unknowns) is an eqn in the form
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$$ a_1x_1 + a_2x_2 + ... + a_nx_n = b$$
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where $a_1, a_2, ..., a_n, b$ are constants.
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\begin{note}
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In a linear system, we don't assume that $a_1, a_2, ..., a_n$ are not all 0
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\begin{itemize}
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\item If $a_1 = ... = a_n = 0$ but $b \neq 0$, it is \textbf{inconsistent}
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E.g. $0x_1 + 0x_2 = 1$
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\item If $a_1 = ... = a_n = b = 0$, it is a \textbf{zero equation}
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E.g. $0x_1 + 0x_2 = 0$
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\item Linear equation which is not a zero equation is a \textbf{nonzero equation}
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E.g. $2x_1 - 3x_2 = 4$
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\item The following are not linear equations
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\begin{itemize}
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\item $xy = 2$
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\item $\sin\theta + \cos\phi = 0.2$
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\item $x_1^2 + x_2^2 + ... + x_n^2 = 1$
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\item $x = e^y$
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\end{itemize}
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\end{itemize}
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\end{note}
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In the $xyz$ space, linear equation $ax + by + cz = d$ where $a, b, c > 0$ represents a plane
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\subsubsection{Solutions to a Linear Equation}
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Let $a_1x_1 + a_2x_2 + ... + a_nx_n = b$ be a linear eqn in n variables \\
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For real numbers $s_1+ s_2+ ... + s_n$, if $a_1s_1 + a_2s_2 + ... + a_ns_n = b$, then $x_1 = s_1, x_2 = s_2, x_n = s_n$ is a solution to the linear equation \\
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The set of all solutions is the \textbf{solution set}\\
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Expression that gives the entire solution set is the \textbf{general solution}
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\textbf{Zero Equation} is satified by any values of $x_1, x_2,... x_n$
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General solution is given by $(x_1, x_2, ..., x_n) = (t_1, t_2, ..., t_n)$
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\subsubsection{Examples: Linear equation $4x-2y = 1$}
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\begin{itemize}
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\item x can take any arbitary value, say t
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\item $x = t \Rightarrow y = 2t - \frac{1}{2}$
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\item General Solution:
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$
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\begin{cases}
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x = t & \text{t is a parameter}\\
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y = 2t - \frac{1}{2}
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\end{cases}
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$
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\item y can take any arbitary value, say s
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\item $y = s \Rightarrow x = \frac{1}{2}s + \frac{1}{4}$
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\item General Solution:
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$
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\begin{cases}
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y = s & \text{s is a parameter}\\
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x = \frac{1}{2}s + \frac{1}{4}
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\end{cases}
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$
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\end{itemize}
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\subsubsection{Example: Linear equation $x_1 - 4x_2 + 7x_3 = 5$}
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\begin{itemize}
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\item $x_2$ and $x_3$ can be chosen arbitarily, $s$ and $t$
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\item $x_1 = 5 + 4s -7t$
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\item General Solution:
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$
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\begin{cases}
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x_1 = 5 + 4s -7t \\
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x_2 = s & s, t \text{ are arbitrary parameters}\\
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x_3 = t \\
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\end{cases}
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$
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\end{itemize}
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\subsection{Linear System}
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Linear System of m linear equations in n variables is
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\begin{equation}
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\begin{cases}
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a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n = b_1 \\
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a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n = b_2 \\
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\vdots \\
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a_{m1}x_1 + a_{m2}x_2 + ... + a_{mn}x_n = b_m \\
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\end{cases}
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\end{equation}
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where $a_{ij}, b$ are real constants and $a_{ij}$ is the coeff of $x_j$ in the $i$th equation
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\begin{note} Linear Systems
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\begin{itemize}
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\item If $a_{ij}$ and $b_i$ are zero, linear system is called a \textbf{zero system}
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\item If $a_{ij}$ and $b_i$ is nonzero, linear system is called a \textbf{nonzero system}
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\item If $x_1 = s_1, x_2 = s_2, ..., x_n = s_n$ is a solution to \textbf{every equation} in the system, then its a solution to the system
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\item If every equation has a solution, there might not be a solution to the system
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\item \textbf{Consistent} if it has at least 1 solution
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\item \textbf{Inconsistent} if it has no solutions
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\end{itemize}
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\end{note}
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\subsubsection{Example}
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\begin{equation}
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\begin{cases}
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a_1x + b1_y = c_1 \\
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a_2x + b2_y = c_2 \\
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\end{cases}
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\end{equation}
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where $a_1, b_1, a_2, b_2$ not all zero
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In $xy$ plane, each equation represents a straight line, $L_1, L_2$
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\begin{itemize}
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\item If $L_1, L_2$ are parallel, there is no solution
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\item If $L_1, L_2$ are not parallel, there is 1 solution
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\item If $L_1, L_2$ coinside(same line), there are infinitely many solution
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\end{itemize}
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\begin{equation}
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\begin{cases}
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a_1x + b1_y + c_1z = d_1 \\
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a_2x + b2_y + c_2z = d_2 \\
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\end{cases}
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\end{equation}
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where $a_1, b_1, c_1, a_2, b_2, c_2$ not all zero
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In $xyz$ space, each equation represents a plane, $P_1, P_2$
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\begin{itemize}
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\item If $P_1, P_2$ are parallel, there is no solution
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\item If $P_1, P_2$ are not parallel, there is $\infty$ solutions (on the straight line intersection)
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\item If $P_1, P_2$ coinside(same plane), there are infinitely many solutions
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\item Same Plane $\Leftrightarrow a_1 : a_2 = b_1 : b_2 = c_1 : c_2 = d_1: d_2$
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\item Parallel Plane $\Leftrightarrow a_1 : a_2 = b_1 : b_2 = c_1 : c_2$
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\item Intersect Plane $\Leftrightarrow a_1 : a_2, b_1 : b_2, c_1 : c_2$ are not the same
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\end{itemize}
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\subsection{Augmented Matrix}
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$
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\begin{amatrix}{3}
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a_{11} & a_{12} & a_{1n} & b_1 \\
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a_{21} & a_{12} & a_{2n} & b_2 \\
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a_{m1} & a_{m2} & a_{mn} & b_m \\
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\end{amatrix}
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$
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\subsection{Elementary Row Operations}
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To solve a linear system we perform operations:
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\begin{itemize}
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\item Multiply equation by nonzero constant
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\item Interchange 2 equations
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\item add a constant multiple of an equation to another
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\end{itemize}
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Likewise, for a augmented matrix, the operations are on the \textbf{rows} of the augmented matrix
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\begin{itemize}
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\item Multiply row by nonzero constant
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\item Interchange 2 rows
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\item add a constant multiple of a row to another row
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\end{itemize}
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\subsection{Recap}
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Given the linear equation $a_1x_1 + a_2x_2 + ... + a_nx_n = b$
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\begin{enumerate}
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\item $a_1 = a_2 = ... = a_n = b = 0$ zero equation
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Solution: $x_1 = t_1, x_2 = t_2, ... = x_n = t_n$
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\item $a_1 = a_2 = ... = a_n = 0 \neq b$ inconsistent
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No Solution
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\item Not all $a_1 ... a_n$ are zero.
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Set $n-1$ of $x_i$ as params, solve for last variable
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\end{enumerate}
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\subsection{Elementary Row Operations Example}
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\begin{center}
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\begin{minipage}{.3\linewidth}
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\systeme{x+y+3z = 0, 2x-2y+2z=4, 3x+9y=3}
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\end{minipage}%
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\begin{minipage}{.3\linewidth}
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\begin{equation*}
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\begin{amatrix}{3}
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1 & 1 & 3 & 0 \\
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2 & 2 & 2 & 4 \\
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3 & 9 & 0 & 3
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\end{amatrix}
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\end{equation*}
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\end{minipage}
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\end{center}
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\subsection{Row Equivalent Matrices}
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2 Augmented Matrices are row equivalent if one can be obtained from the other by a series of elementary row operations
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Given a augmented matrix $A$, how to find a row equivalent augmented matrix B of which is of a \textbf{simple} form?
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\subsection{Row Echelon Form}
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\begin{defn}[Row Echelon Form (Simple)]
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Augmented Matrix is in row-echelon form if
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\begin{itemize}
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\item Zero rows are grouped together at the bottom
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\item For any 2 successive nonzero rows, The first nonzero number in the lower row appears to the right of the first nonzero number on the higher row
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$\begin{amatrix}{4}
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0 & 0 & 1 & 2 & 3 \\
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0 & 0 & 0 & 1 & 2 \\
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\end{amatrix}$
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\item Leading entry if a nonzero row is a \textbf{pivot point}
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\item Column of augmented matrix is called
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\begin{itemize}
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\item \textbf{Pivot Column} if it contains a pivot point
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\item \textbf{Non Pivot Column} if it contains no pivot point
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\end{itemize}
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\item Pivot Column contains exactly 1 pivot point
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\# of pivots = \# of leading entries = \# of nonzero rows
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\end{itemize}
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\end{defn}
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Examples of row echlon form:
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\begin{equation*}
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\begin{amatrix}{2}
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3 & 2 & 1 \\
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\end{amatrix}
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\begin{amatrix}{2}
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1 & -1 & 0 \\
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0 & 1 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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2 & 1 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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1 & 2 & 3 & 4 \\
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0 & 1 & 1 & 2 \\
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0 & 0 & 2 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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0 & 1 & 2 & 8 & 1 \\
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0 & 0 & 0 & 0 & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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Examples of NON row echlon form:
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\begin{equation*}
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\begin{amatrix}{2}
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0 & \textbf{1} & 0 \\
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\textbf{1} & 0 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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0 & 0 & \textbf{1} \\
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\textbf{1} & -1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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\textbf{1} & 0 & 2 & 1 \\
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0 & \textbf{1} & 0 & 2 \\
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0 & \textbf{1} & 1 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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\textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} \\
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1 & 0 & 2 & 0 & 1 \\
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0 & 0 & 0 & 1 & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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\subsection{Reduced Row-Echelon Form}
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\begin{defn}[Reduced Row-Echelon Form]
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\label{def:rref}
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Suppose an augmented matrix is in row-echelon form. It is in \textbf{reduced row-echelon form} if
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\begin{itemize}
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\item Leading entry of every nonzero row is 1
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\subitem Every pivot point is one
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\item In each pivot column, except the pivot point, all other entries are 0.
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\end{itemize}
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\end{defn}
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Examples of reduced row-echelon form:
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\begin{equation*}
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\begin{amatrix}{2}
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1 & 2 & 3 \\
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\end{amatrix}
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\begin{amatrix}{2}
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0 & 0 & 0 \\
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0 & 0 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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1 & 0 & 0 & 1 \\
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0 & 1 & 0 & 2 \\
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0 & 0 & 1 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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0 & 1 & 2 & 0 & 1 \\
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0 & 0 & 0 & 1 & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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Examples of row-echelon form but not reduced: (pivot point is not 1 / all other elements \textbf{in pivot column} must be zero)
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\begin{equation*}
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\begin{amatrix}{2}
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\textbf{3} & 2 & 1 \\
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\end{amatrix}
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\begin{amatrix}{2}
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1 & \textbf{-1} & 0 \\
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0 & 1 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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\textbf{2} & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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\textbf{-1} & 2 & 3 & 4 \\
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0 & 1 & 1 & 2 \\
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0 & 0 & 2 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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0 & 1 & 2 & \textbf{8} & 1 \\
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0 & 0 & 0 & \textbf{4} & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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To note: 2nd matrix has -1 in the pivot column, but 5th matrix has 2 in a non-pivot column so its fine
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\subsection{Solving Linear System}
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If Augmented Matrix is in reduced row-echelon form, then solving it is easy
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\begin{equation*}
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\begin{amatrix}{3}
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1 & 0 & 0 & 1 \\
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0 & 1 & 0 & 2 \\
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0 & 0 & 1 & 3 \\
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\end{amatrix}
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\text{then } x_1 = 1, x_2 = 2, x_3 = 3
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\end{equation*}
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\begin{note}
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\begin{itemize}
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\item If any equations in the system is inconsistent, the whole system is inconsistent
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\end{itemize}
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\end{note}
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\subsubsection{Examples}
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Augmented Matrix: $\begin{amatrix}{4}
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1 & -1 & 0 & 3 & -2 \\
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0 & 0 & 1 & 2 & 5 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}$
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\begin{itemize}
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\item The zero row can be ignored.
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\systeme{x_1 - x_2 + 3x_4 = -2, x_3 + 2x_4 = 5}
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\item Degree of freedom(\# cols): 4, number of restrictions (\# pivot cols): 2, arbitrary vars(\# non pivot cols): 4-2 = 2. Set this to the non-pivot cols
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\end{itemize}
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\begin{enumerate}
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\item Let $x_4 = t$ and sub into 2nd eqn
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\subitem $x_3 + 2t = 5 \Rightarrow x_3 = 5-2t$
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\item sub $x_4 = t$ into 1st eqn
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\subitem $x_1 - x_2 + 3t = -2$
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\subitem Let $x_2 = s$. Then $x_1 = -2 + s - 3t$
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\item Infinitely many sols with ($s$ and $t$ as arbitrary params)
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\subitem $x_1 = -2 + s - 3t, x_2 = s, x_3 = 5-2t, x_4 = t$
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\end{enumerate}
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Augmented Matrix: $\begin{amatrix}{5}
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0 & 2 & 2 & 1 & -2 & 2 \\
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0 & 0 & 1 & 1 & 1 & 3 \\
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0 & 0 & 0 & 0 & 2 & 4 \\
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\end{amatrix}$
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\begin{itemize}
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\item \systeme{0x_1 + 2x_2 + 2x_3 + 1x_4 -2x_5 = 2,x_3 + x_4 +x_5 = 3,2x_5 = 4}
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\item Degree of freedom: 5, number of restrictions: 3, arbitrary vars: 5-3 = 2
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\end{itemize}
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\begin{enumerate}
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\item by 3rd eqn, $2x_5 = 4 \Rightarrow x_5 = 2$
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\item sub $x_5 = 2$ into 2nd eqn
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\subitem $x_3 + x_4 + 2 = 3 \Rightarrow x_3 + x_4 = 1$
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\subitem let $x_4 = t$. Then $x_3 = 1-t$
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\item sub $x_5 = 2, x_3 = 1-t, x_4 = t$ into 1st eqn
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\subitem $2x_2 + 2(1-t) + t - 2(2) = 2 \Rightarrow 2x_2 -t = 4 \Rightarrow x_2 = \frac{t}{2} + 2$
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\item system has inf many solns: $x_1 = s, x_2 = \frac{t}{2} + 2, x_3 = 1-t, x_4 = t, x_5 = 2$ where $s$ and $t$ are arbitrary
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\end{enumerate}
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\subsubsection{Algorithm}
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Given the augmented matrix is in row-echelon form.
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\begin{enumerate}
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\item Set variables corresponding to non-pivot columns to be arbitrary parameters
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\item Solve variables corresponding to pivot columns by back substitution (from last eqn to first)
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\end{enumerate}
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\subsection{Gaussian Eliminiation}
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\begin{defn}[Gaussian Elimination]\ \\
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\begin{enumerate}
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\item Find the left most column which is not entirely zero
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\item If top entry of such column is 0, replace with nonzero number by swapping rows
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\item For each row below top row, add multiple of top row so that leading entry becomes 0
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\item Cover top row and repeat to remaining matrix
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\end{enumerate}
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\end{defn}
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\begin{note}[Algorithm with Example]\ \\
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$\begin{amatrix}{6}
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0 & 0 & 0 & 2 & 4 & 2 & 8 \\
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0 & 1 & 2 & 4 & 5 & 3 &-9 \\
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0 &-2 &-4 &-5 &-4 & 3 & 6 \\
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\end{amatrix}$
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\begin{enumerate}
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\item Find the left most column which is not all zero (2nd column)
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\item Check top entry of the selection. If its zero, replace it by a nonzero number by interchanging the top row with another row below
|
|
\subitem $\begin{amatrix}{6}
|
|
0 & 1 & 2 & 4 & 5 & 3 &-9 \\
|
|
0 & 0 & 0 & 2 & 4 & 2 & 8 \\
|
|
0 &-2 &-4 &-5 &-4 & 3 & 6 \\
|
|
\end{amatrix}$
|
|
\item For each row below the top row, adda suitable multiple of top row so that leading entry becomes 0.
|
|
\subitem $2R_1 + R_3$ will ensure that the -2 turns to 0
|
|
\subitem $\begin{amatrix}{6}
|
|
0 & 1 & 2 & 4 & 5 & 3 &-9 \\
|
|
0 & 0 & 0 & 2 & 4 & 2 & 8 \\
|
|
0 & 0 & 0 & 3 & 6 & 9 &-12\\
|
|
\end{amatrix}$
|
|
\item Cover top row and repeat procedure to the remaining matrix
|
|
\subitem $\begin{amatrix}{6}
|
|
0 & 1 & 2 & 4 & 5 & 3 &-9 \\
|
|
\hline
|
|
0 & 0 & 0 & 2 & 4 & 2 & 8 \\
|
|
0 & 0 & 0 & 3 & 6 & 9 &-12\\
|
|
\end{amatrix}$
|
|
\subitem Look at $C_4$. $R_3 \times -1.5R_2$ will set $R_3C_4$ to zero.
|
|
\subitem $\begin{amatrix}{6}
|
|
0 & 1 & 2 & 4 & 5 & 3 &-9 \\
|
|
\hline
|
|
0 & 0 & 0 & 2 & 4 & 2 & 8 \\
|
|
0 & 0 & 0 & 0 & 0 & 6 & -24\\
|
|
\end{amatrix}$
|
|
\subitem This is now in row echelon form.
|
|
\end{enumerate}
|
|
Only use $R_i \Leftrightarrow R_j or R_i + CR_j$ in this method.
|
|
\end{note}
|
|
|
|
\subsection{Gauss-Jordan Elimination}
|
|
\begin{defn}[Gauss Joran Elimination]\ \\
|
|
\begin{enumerate}
|
|
\item[1-4.] Use Gaussian Eliminiation to get row-echelon form
|
|
\setcounter{enumi}{4}
|
|
\item For each nonzero row, multiply a suitable constant so pivot point becomes 1
|
|
\item Begin with last nonzero row and work backwords
|
|
\subitem Add suitable multiple of each row to the rows above to introduce 0 above pivot point
|
|
\end{enumerate}
|
|
\begin{itemize}
|
|
\item Every matrix has a unique reduced row-echelon form.
|
|
\item Every nonzero matrix has infinitely many row-echelon ofrm
|
|
\end{itemize}
|
|
\end{defn}
|
|
\begin{note}[Gauss Jordan Elimination Example] Suppose an augmented matrix is in row-echelon form.
|
|
$\begin{amatrix}{5}
|
|
1 & 2 & 4 & 5 & 3 & -9 \\
|
|
0 & 0 & 2 & 4 & 2 & 8 \\
|
|
0 & 0 & 0 & 0 & 6 & -24 \\
|
|
\end{amatrix}$
|
|
\begin{enumerate}
|
|
\item All pivot points must be 1
|
|
\subitem multiply $R_2$ by $\frac{1}{2}$ and $R_3$ by $\frac{1}{6}$
|
|
\subitem $\begin{amatrix}{5}
|
|
1 & 2 & 4 & 5 & 3 & -9 \\
|
|
0 & 0 & 1 & 2 & 1 & 4 \\
|
|
0 & 0 & 0 & 0 & 1 & -4 \\
|
|
\end{amatrix}$
|
|
\item In each pivot col, all entries other than pivot point must be 0. Work backwards
|
|
\subitem $R_1 + -3R_1$, $R_2 + -R_1$
|
|
\subitem $\begin{amatrix}{5}
|
|
1 & 2 & 4 & 5 & 0 & 3 \\
|
|
0 & 0 & 1 & 2 & 0 & 8 \\
|
|
0 & 0 & 0 & 0 & 1 & -4 \\
|
|
\end{amatrix}$
|
|
\subitem $R_1 + -4R_2$
|
|
\subitem $\begin{amatrix}{5}
|
|
1 & 2 & 0 & -3 & 0 & -29 \\
|
|
0 & 0 & 1 & 2 & 0 & 8 \\
|
|
0 & 0 & 0 & 0 & 1 & -4 \\
|
|
\end{amatrix}$
|
|
|
|
\end{enumerate}
|
|
\end{note}
|
|
|
|
\subsection{Review}
|
|
|
|
\begin{align*}
|
|
I: & cR_i, c \neq 0 \\
|
|
II: & R_i \Leftrightarrow R_j \\\
|
|
III: & R_i \Rightarrow R_i + cR_j
|
|
\end{align*}
|
|
|
|
Solving REF:
|
|
\begin{enumerate}
|
|
\item Set var -> non-pivot cols as params
|
|
\item Solve var -> pivot cols backwards
|
|
|
|
\# of nonzero rows = \# pivot pts = \# of pivot cols
|
|
\end{enumerate}
|
|
|
|
Gaussian Elimination
|
|
\begin{enumerate}
|
|
\item Given a matrix $A$, find left most non-zero \textbf{column}. If the leading number is NOT zero, use $II$ to swap rows.
|
|
\item Ensure the rest of the column is 0 (by subtracting the current row from tht other rows)
|
|
\item Cover the top row and continue for next rows
|
|
\end{enumerate}
|
|
|
|
\subsection{Consistency}
|
|
\begin{defn}[Consistency]\ \\
|
|
Suppose that $A$ is the Augmented Matrix of a linear system, and $R$ is a row-echelon form of $A$.
|
|
\begin{itemize}
|
|
\item When the system has no solution(inconsistent)?
|
|
\subitem There is a row in $R$ with the form $(0 0 ... 0 | \otimes)$ where $\otimes \neq 0$
|
|
\subitem Or, the last column is a pivot column
|
|
\item When the system has exactly one solution?
|
|
\subitem Last column is non-pivot
|
|
\subitem All other columns are pivot columns
|
|
\item When the system has infinitely many solutions?
|
|
\subitem Last column is non-pivot
|
|
\subitem Some other columns are non-pivot columns.
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\begin{note} Notations\ \\
|
|
For elementary row operations
|
|
\begin{itemize}
|
|
\item Multiply $i$th row by (nonzero) const $k$: $kR_i$
|
|
\item Interchange $i$th and $j$th rows: $R_i \leftrightarrow R_j$
|
|
\item Add $K$ times $i$th row to $j$th row: $R_j + kR_i$
|
|
\end{itemize}
|
|
\textbf{Note}
|
|
\begin{itemize}
|
|
\item $R_1 + R_2$ means "add 2nd row to the 1st row".
|
|
\item $R_2 + R_1$ means "add 1nd row to the 2st row".
|
|
\end{itemize}
|
|
|
|
\textbf{Example}
|
|
$$ \begin{pmatrix} a \\ b \end{pmatrix}
|
|
\xrightarrow{R_1 + R_2} \begin{pmatrix} a + b \\ b \end{pmatrix}
|
|
\xrightarrow{R_2 + (-1)R_1} \begin{pmatrix} a + b \\ -a \end{pmatrix}
|
|
\xrightarrow{R_1 + R_2} \begin{pmatrix} b \\ -a \end{pmatrix}
|
|
\xrightarrow{(-1)R_2}\begin{pmatrix} b \\ a \end{pmatrix}
|
|
$$
|
|
\end{note}
|
|
|
|
\subsection{Homogeneous Linear System}
|
|
|
|
\begin{defn}[Homogeneous Linear Equation \& System]\ where
|
|
\begin{itemize}
|
|
\item Homogeneous Linear Equation: $a_1x_1 + a_2x_2 + ... + a_nx_n = 0 \iff x_1 = 0, x_2 = 0,... , x_n = 0$
|
|
\item Homogeneous Linear Equation: $\begin{cases}
|
|
a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n = 0 \\
|
|
a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n = 0 \\
|
|
\vdots \\
|
|
a_{m1}x_1 + a_{m2}x_2 + ... + a_{mn}x_n = 0 \\
|
|
\end{cases}$
|
|
|
|
\item This is the trivial solution of a homogeneous linear system.
|
|
\end{itemize}
|
|
|
|
You can use this to solve problems like Find the equation $ax^2 + by^2 + cz^2 = d$, in the $xyz$ plane which contains the points $(1, 1, -1), (1, 3, 3), (-2, 0, 2)$.
|
|
|
|
\begin{itemize}
|
|
\item Solve by first converting to Augmented Matrix, where the last column is all 0. During working steps, this column can be omitted.
|
|
\item With the \hyperref[def:rref]{RREF}, you can set $d$ as $t$ and get values for $a, b, c$ in terms of $t$.
|
|
\item sub in $t$ into the original equation and factorize $t$ out from both sides, for values where $t \neq 0$
|
|
\end{itemize}
|
|
|
|
\end{defn}
|