~~REVEAL~~

==== Another look at the last example ====
  * $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
  * Find solutions of this [[system of linear equations|system]] by applying operations
  * Aim to end up with a very simple sort of system where we can see the solutions easily.

==== ====
  * $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
  * Replace equation (2) with $(2)-2\times (1)$:
  * $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
  * Now replace equation (1) with $(1)-3\times (2)$
  * $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$

==== ====
  * $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
  * can easily rearrange (1) to find $x$ in terms of $z$
  * can easily rearrange (2) to find $y$ in terms of $z$ 
  * Since $z$ can take any value, write $z=t$ where $t$ is a "free parameter" 
  * (which means $t$ can be any real number, or $t\in \mathbb{R}$).
  * Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*}

==== ====
  * Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*}
  * Can also write this in "vector form":
  * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.$
  * This is the equation of the line where the two planes described by the original equations intersect.

==== ====
  * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}$
  * For each value of $t$, we get a different solution (a different point on the line of intersection). 
  * e.g. take $t=0$ to see that $(-16,7,0)$ is a solution
  * take $t=1.5$ to see that $(-16+1.5\times 5,7+1.5\times (-2),1.5) = (-8.5,4,1.5)$ is another solution
  * etc.
  * This works for any value $t\in\mathbb{R}$, and every solution may be written in this way.

==== Observations ====

  - The operations we applied to the original linear system don't change the set of solutions. This is because each operation is reversible.
  - Writing out the variables $x,y,z$ each time is unnecessary:

  * erase the variables from the system $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$$ 
  * write all the numbers in a grid, or a **matrix**
  * we get $\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}$

==== ====
  * System of linear equations: $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
  * $\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}$ is called the **augmented matrix** of this linear system
  * Each **row** corresponds to one **equation**.
  * Each **column** corresponds to one **variable**
    * (except the last column, which has the right-hand-sides of the equations)
  * Instead of performing operations on equations, **we can perform operations on the rows of this matrix**.

==== ====
\begin{align*} 
\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix} 
&\xrightarrow{R2\to R2-2\times R1}
\begin{bmatrix} 1&3&1&5\\0&1&2&7\end{bmatrix} 
\\[6pt]&\xrightarrow{R1\to R1-3\times R1}
\begin{bmatrix} 1&0&-5&-16\\0&1&2&7\end{bmatrix}
\end{align*}
  * Translate back into equations and solve:
  * $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
  * $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.$

==== ====

This method always works: 
  * take any system of linear equations
  * write down a corresponding matrix (the **augmented matrix**)
  * perform reversible operations on the rows of this matrix to get a "nicer" matrix
  * write down a new system of linear equations with the same solutions as the original system. 
  * Hopefully the new system will be easy to solve...
  * and the solutions haven't changed, so we'll have solved the original system!


===== The augmented matrix and elementary operations =====

==== Definition ====

{{page>augmented matrix}}

The numbers in this matrix are called its **entries**. 

==== Example ====
  * Find the augmented matrix of the linear system\begin{align*}3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}
  * We can rewrite it as \begin{align*}3x+4y+7z&=2\\{\color{red}1}x+{\color{red}0y}+3z&=0\\{\color{red}0x}+{\color{red}1}y-2z&=5\end{align*}
  * So the augmented matrix is\[ \begin{bmatrix} 3&4&7&2\\1&0&3&0\\0&1&-2&5\end{bmatrix}.\]

==== Elementary operations on a system of linear equations ====

{{page>elementary operations on a linear system for slides}}

==== Why do elementary operations leave the solutions of systems unchanged? ====

  * We do the same thing to the left hand side and the right hand side of each equation...
    * so any solution to the original system will also be a solution to the new system.
  * These operations are all reversible (using operations of the same type)...
    * so any solution to the new system will also be a solution to the original system.

==== Elementary row operations on a matrix ====

{{page>elementary row operation for slides}}

==== Example ====

Use [[EROs]] to find the intersection of the planes
\begin{align*} 3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}

==== Solution 1 ====

\begin{align*} 
\def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}
\def\ar#1{\\\xrightarrow{#1}&}
&\go{3&4&7&2}{1&0&3&0}{0&1&-2&5}
\ar{\text{reorder rows}}\go{1&0&3&0}{0&1&-2&5}{3&4&7&2}
\ar{R3\to R3-3R1}\go{1&0&3&0}{0&1&-2&5}{0&4&-2&2}
\ar{R3\to R3-4R2}\go{1&0&3&0}{0&1&-2&5}{0&0&6&-18}
\ar{R3\to \tfrac16 R3}\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3}
\end{align*}

==== ====

$\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3}$

  * from the last row, we get $z=-3$
  * from the second row, we get $y-2z=5$
    * so $y-2(-3)=5$
    * so $y=-1$
  * from the first row, we get $x+3z=0$
    * so $x+3(-3)=0$
    * so $x=9$
  * Conclusion: $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}$ is the only solution.

==== Solution 2 ====

We start in the same way, but by performing more [[EROs]] we make the algebra at the end simpler.

\begin{align*} 
\def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}
\def\ar#1{\\\xrightarrow{#1}&}
&\go{3&4&7&2}{1&0&3&0}{0&1&-2&5}
\ar{\ldots \text{same EROs as above}\ldots}\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3}
\ar{R2\to R2+2R3}\go{1&0&3&0}{0&1&0&-1}{0&0&1&-3}
\ar{R1\to R1-3R3}\go{1&0&0&9}{0&1&0&-1}{0&0&1&-3}
\end{align*}

==== ====
\[ \go{1&0&0&9}{0&1&0&-1}{0&0&1&-3}\]

  * from the last row, we get $z=-3$
  * from the second row, we get $y=-1$
  * from the first row, we get $x=9$
  * So $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}$ is the only solution.

==== Discussion ====

Both solutions use EROs to transform the [[augmented matrix]].

  * Solution 1: $\left[\begin{smallmatrix}1&0&3&0\\0&1&-2&5\\0&0&1&-3\end{smallmatrix}\right]$.
    * "Staircase pattern": 1s on "steps", zeros below steps
    * Called **row echelon form** 
    * Needed algebra to finish solution.
  * Solution 2: $\left[\begin{smallmatrix}1&0&0&9\\0&1&0&-1\\0&0&1&-3\end{smallmatrix}\right]$ 
    * Staircase with zeros **above** 1s on steps (and below).
    * Called **reduced row echelon form**
    * No extra algebra needed to finish solution.