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--- lecture_3_slides [2016/02/01 17:21] – rupert
+++ lecture_3_slides [2016/02/01 17:29] (current) – rupert
@@ Line 33: / Line 33: @@
   * 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
+  * 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. If we erase the variables from the system $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$$ and write all the numbers in a grid, or a **matrix**, we get $\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}$
+  - 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 **row** corresponds to one **equation**.
-  * The first column corresponds to the $x$ variable, the second to $y$, the third to $z$
+  * Each **column** corresponds to one **variable**
-  * Numbers in the final column are the right hand sides of the equations.
+    * (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.
+  * Instead of performing operations on equations, **we can perform operations on the rows of this matrix**.
 ==== ====
@@ Line 98: / Line 102: @@
 {{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.