Systems of Linear equations
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Changes of thinking
- in the previous lecture we thought to solve a system of LE by isolatating a variable and plugging it in
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A collection of m linear equations in n variables is called a system
- given a system of m linear equations in n variables we would write
- is a solution to the system if all m linear equations are satisfied when we set
- if there is at least one solution it is called consistent
- if there is no solutions it is called inconsistent
- given a system of m linear equations in n variables we would write
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Elementary operations:
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1) multiplying an equation (row) by a non-zero number$$
- \begin{align} x+3y=2 \ 2x+4y=4 \ -x+y=10 \end{align}
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\begin{align} 2x+4y=4 \x+3y=2 \
-x+y=10 \end{align}
- #### 3) we can add/subtract rows from one another - subtracting row 1 by row 2\begin{align} -x-y=-4 \ 2x+4y=4 \ -x+y=10 \end{align}
- ### Example - Determin wether $\vec{w}=(1,1,0)$ is a linear combination of $\vec{v}_{1}=(0,1,1)$ and $\vec{v}_{2}=(1,0,1)$ - find s,t s.uch that $$ - \begin{align} s\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} +t\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \end{align} $$ - then we can simplify\begin{bmatrix} t \ s \ t+s \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix}
$$
- then we can seperate the letters in their own collumns $$
- \begin{bmatrix}
0 & 1 & |1 \ 1 & 0 & |1 \ 1 & 1 & |0 \end{bmatrix}
- and we can use RREF $$ - \begin{bmatrix} 1 & 0 & |0 \\ 0 & 1 & |0 \\ 0 & 0 & |1 \end{bmatrix} $$ as we can see that 1=0 $\therefore$ it is inconsistent - ## systems with infinitely many solutions - #### example: intersection of two planes in $R^3$ - $$ \begin{align} -x+2y+z=1 \\ x+y-3z=5 \end{align} $$$$ \begin{bmatrix} -1 & 2 & 1 & |1 \\ 1 & 1 & -3 & |5 \end{bmatrix} \to R_{2}+R_{1}\to\begin{bmatrix} -1 & 2 & 1 & |1 \\ 0 & 3 & -2 & |6 \end{bmatrix} $$ - then after solving to REF - $$ \begin{bmatrix} 1 & 0 & -7/3 & |3 \\ 0 & 1 & -2/3 & |2 \end{bmatrix}\implies \begin{align} -x-\frac{7}{3}z=3 \\ y-\frac{2}{3}z=2 \end{align} - let $z=t,$then $x=3+\frac{7}{3}t,y=2+\frac{2}{3}t$
$$
\begin{bmatrix}
x \ y \ z \end{bmatrix}=\begin{bmatrix} 3+7t/3 \ 2+2t/3 \ t \end{bmatrix}(t R)\implies \begin{bmatrix} x \ y \ z \end{bmatrix}=\begin{bmatrix} 3 \ 2 \ 0 \end{bmatrix}+t\begin{bmatrix} 7/3 \ 2/3 \ t \end{bmatrix}