#uw/notes #uw/class/math208 
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- Not all systems of linear equations will be in echelon form and easily solvable
- There are three **elementary operations** that can be used to create a new system that is equivalent to the old one
	- Interchange the position of two equations
	- Multiply an equation by a nonzero constant
	- Add a multiple of one equation to another
	- $\sim$ is used to indicate the transformation between equivalent linear systems
- matrices can be used to simplify when working with systems of linear equations
	- matrices with all constant terms of a linear system of equations are called a **augmented matrix**
	- $$\begin{gathered} \text{Linear System} \\ a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} = b_{2} \\ a_{31}x_{1} + a_{32}x_{2} + a_{33}x_{3} = b_{3} \end{gathered} \quad \sim \quad \begin{gathered} \text{Augmented Matrix} \\ \left[\begin{array}{ccc|c}
	a_{11} & a_{12} & a_{13} & b_{1} \\ a_{21} & a_{22} & a_{23} & b_{2} \\ a_{31} & a_{32} & a_{33} & b_{3} 
	\end{array}\right] \end{gathered}$$
	- Same elementary operations can be used with augmented matrices, now it is with rows instead of equations.
- Gaussian Elimination
	- this is converting a matrix to **echelon form** (or row echelon form)
		- this is when every leading term is a column to the left of the one below it and any zero rows are at the bottom
	- Very important method for many applications, used to solve systems of linear equations
	- once the augmented matrix is in echelon form it can be converted back into a linear system of equations (that is not in echelon form) and solved
	- the pivot is the coefficient of the leading terms (or the free variables)
		- The first non-zero term in a row
- Gauss-Jordan elimination
	- this can make it easier to find the general solution of the system
	- 1. multiply each nonzero row by the inverse of the pivot so every pivot is 1
	- 2. manipulate so every pivot only has zeros above it
	- This means that leading variables are only in the equations that they lead
	- This puts the matrix in **reduced echelon form**
- Homogeneous linear system
	- when the $b_{n}$ term of each equation is zero
	- There is a "trivial solution" where each variable equals zero
	- There can also be non-trivial solutions 


To find when the system $$\begin{gathered} 3x_{1} + 6x_{2} = 2 \\ hx_{1} + 12x_{2} = k \end{gathered}$$ has no solutions
$$\left[\begin{array}{cc|c}
3 & 6 & 2 \\ h & 12 & k
\end{array}\right] \sim \left[\begin{array}{cc|c}
h & 2h & \frac{2h}{3} \\ h & 12 & k
\end{array}\right] \sim \left[\begin{array}{cc|c}
h & 2h & \frac{2h}{3} \\ 0 & 12 - 2h & k - \frac{2h}{3}
\end{array}\right] $$
There are no solutions the system of equations when there is no pivot in a row and the value on the right is not equal to zero (if the value on the right is also equal to zero than that equation is essentially $0=0$ which is always true).
This means that there are no solutions when $$\begin{gather} 12-2h = 0 \\[1.5ex] h=6 \end{gather}$$ and $$ \begin{gather} k-\frac{2h}{3} \ne 0 \\[1.5ex]  k-4 \ne 0 \\[1.5ex]  k \ne 4 \end{gather}$$