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1.1,1.2 Systems of Linear Equations Row Reduction and Echelon Forms…

- - - - Use the \(x_1\) term in the first equation of a system to eliminate the \(x_1\) terms in the other equations. THen use the \(x_2\) term in the second equation to eliminate the \(x_2\) terms in the other equations and so on
        
        The 3 basic equations we use to simplify a system are:
        
        Replace one equation by the sum of itself and a multiple of another equation
        
        interchange two equations
        
        multiply all the terms in an equation by a nonzero constant
    - - Once we have all the x variables plug it back into the original system
        
        If done right the left side should equal the right side
        
        The right side being b
- - - - In 3D the equations represent planes so if there's one solution it would look like
  - - - A system is consistent if it has at least one solution
      - Not consistent if it has no solution
    - - A system is consistent if a row of an augmented matrix does not have the form { 0 ... 0 b }. This means no row has the form 0=b where b is a nonzero matrix
        
        If consistent it can have one solution or an infinite number of solutions
        
        There is only one solution if there are no free variables
        
        An infinite number of solutions if there's a free variable present
  - - - So if we retrace our steps the final matrix we got can be reformed back to it's original form
        
        This mean
        
        Another property was being row equivalent
        
        So if we apply row operations to our final matrix and get our original it means the two linear systems are row equivalent and therefore have the same solution set
- - - - So a 2 by 3 means a matrix with 2 rows and 3 columns
        \( \begin{pmatrix} 1 & 0 & 2\\ 0 & 1 & 3\\ \end{pmatrix} \)
- - - - \( \begin{pmatrix} 1 & 1 & 1\\ 2 & 2 & 2\\ 3 & 9 & 14 \\ \end{pmatrix} \)
      - The first two rows are row equivalent because if we multiply the first equation by 2 or divide the second equation by 2 we get the same equation for both
  - - - It's a nonzero number that's needed to create zeros via row operations
  - - - We can define \(x_3\) to be anything but note by doing so we'll have a different solution set for each specific \(x_3\)
    - - This makes \(x_1\) and \(x_2\) basic variables
  - - - Span is looking at all the possible linear combinations of two vectors
        
        For the most part if there are two vectors then with those vectors we are able to reach any point in 2D by scaling and adding them
        
        In 3D this results in a plane
        
        Let's say we scaled u by 3 then added 2v to it. We'd land in a specific point
        
        Span just means doing this an infinite amount of times in as many variations as possible which is why we get a plane
- - - - Row Echelon Form(R.E.F)
        
        All the leading entries have zeros above them #
        
        This usually results in a triangle of zeros in the bottom left corner
      - Reduced Row Echelon Form(R.R.E.F)
        
        All the leading entries have zeros above and below them
  - - - However, the reduced echelon form ones obtains from a matrix is unique
  - - - Begin with the leftmost nonzero column. This is a pivot column. The pivot position is at the top
        
        Select a nonzero entry in the pivot column as a pivot. If necessary, interchange rows to move this entry into the pivot position
        
        Use row replacement operation to create zeros in all positions below the pivot
        
        Cover (or ignore) the row containing the pivot position and covers all rows, if any, above it. Apply steps 1-3 to the submatrix that remain. Repeat the process until there are no more nonzero rows to modify.
        
        For reduced echelon form
        
        Beginning with the rightmost pivot and working upward and to the left, create zeros above each pivot. If a pivot is not 1, make it 1 by a scaling operation
        
        1 more item...
- - - - Same as
        
        Means \(x_1\)=3 and \(x_2\)=2
      - If we just find the solution set for \( \begin{pmatrix} 1 & 2 & 7\\ -2 & 5 & 4\\ -5 & 6 & -3 \\ \end{pmatrix} \)
        we could also find the answer
    - - u and w are a linear combination of \(v_1\) and \(v_2\)
        
        This means some combination of \(v_1\) and \(v_2\) gives us u and v.
        
        In laymen terms, we know where w lands. What multiple of \(v_1\) and \(v_2\) got us here
  - - - This just means if we do the reduced echelon form or reduced echelon form of the vectors we'll determine what scalars would match
  - - - \(v_1\) is a multiple of \(v_2\)
        
        This means Span{\(v_1\),\(v_2\)} is just a line
        
        Till where does this line extend?
        It goes from 0 to \(v_1\)
        
        We say it goes from 0 to \(v_1\) since \(v_2\) is contained inside \(v_1\)
- - - - If the augmented matrices of two linear systems are row equivalent, then the two
        systems have the same solution set.
        
        We can end up with a different echelon matrix due to different operations taken but solving for the solution gets you the same result
        
        This is why we ca
  - - - Inconsistent
        
        \( \begin{pmatrix} 2 & -3 &2 & 1\\ 4&-8&12&1\\ 0&1&-4&8\\ \end{pmatrix} \)
        
        \( \begin{pmatrix} 2 & -3 &2 & 1\\ 0&-2&8&-1\\ 0&1&-4&8\\ \end{pmatrix} \)
        
        \( \begin{pmatrix} 2 & -3 &2 & 1\\ 0&1&-4&8\\ 0&-2&8&-1\\ \end{pmatrix} \)
        
        \( \begin{pmatrix} 2 & -3 &2 & 1\\ 0&1&-4&8\\ 0&0&0&15\\ \end{pmatrix} \)
        
        1 more item...
      - Consistent
        
        \( \begin{pmatrix} 5 & -3 &3&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix} \)
        
        The rightmost column IS NOT a pivot column since they are all zero terms and pivots are only nonzero
    - - \( \begin{pmatrix} 1 & h &3\\ 3&6&k\\ \end{pmatrix} \)
        
        \( \begin{pmatrix} 1 & h &3\\ 0&6-3h&k-9\\ \end{pmatrix} \)
        
        Where there are at least one free variable, meaning one row must not have a pivot columns
        
        This would be the case where one row has all zeros or if there are more variables than equations
        
        So we would want the last row to be all zeros since there are two variables and two unknowns
        
        1 more item...
        
        No solutions
        
        A system has no solutions when the last column is a pivot column. In other words we're looking for a point where a row is 0 0 0 0 0 b
        where b is nonzero
        
        For this to happen
        6-3h must equal 0
        k-9 must not be equal to zero
        
        1 more item...
        
        One solution
        
        Where there are no free variables, meaning each row must have a pivot columns
        
        This means we want both 6-3h and k-9 to not be equal to 0
        
        1 more item...
- - - - Remember sets like this represent solutions with infinite solutions since there's a free variable