ObsidianVault/College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 - Linear Systems and Matricies.md

#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}