2.1 Systems of Linear Equations

Systems of Linear equations

• Changes of thinking

  • in the previous lecture we thought to solve a system of LE by isolating a variable and plugging it in

• 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
    • $\vec{S}=(s_1,s_2,s_3,\dots ,s_n)$ is a solution to the system if all m linear equations are satisfied when we set $x_1=s_1,\dots ,x_n=s_n$
  • if there is at least one solution it is called consistent
  • if there is no solutions it is called inconsistent

• Elementary operations:

  • 1) multiplying an equation (row) by a non-zero number$$

  • \begin{align} x+3y=2 \
    2x+4y=4 \
    -x+y=10
    \end{align}
  $$-thenwecanmultiplythefirstrowby4$$

\begin{align} 4x+12y=8 \\ 2x+4y=4 \\ -x+y=10 \end{align} $$ - #### 2) we can switch rows around - switching row 2 and 1

\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 - Determine whether $\vec{w}=(1,1,0)$ is a linear combination of $\vec{v}_{1}=(0,1,1)$ and $\vec{v}_{2}=(1,0,1)$ - find s,t such 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}

$$
- then we can separate the letters in their own columns $$
- \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}