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--- lecture_3 [2015/01/27 12:21] – [The augmented matrix of a linear system] rupert
+++ lecture_3 [2016/02/02 10:05] (current) – rupert
@@ Line 1: / Line 1: @@
 Let's look at the example from the end of [[Lecture 2]] more closely:
 $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$$
-We find the solutions of this [[system of linear equations|system]] by apply operations to the system to make a new system, aiming to end up with a very simple sort of system where we can see the solutions easily.
+We find the solutions of this [[system of linear equations|system]] by applying operations to the system to make a new system, aiming to end up with a very simple sort of system where we can see the solutions easily.
 First replace equation (2) with $(2)-2\times (1)$. We'll call the resulting equations (1) and (2) again, although of course we end up with a different system of linear equations:
@@ Line 21: / Line 21: @@
 \[ \begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}\]
 Notice that the first column corresponds to the $x$ variable, the second to $y$, the third to $z$ and the numbers in the final column are the right hand sides of the equations. Each row corresponds to one equation. So instead of performing operations on equations, we can perform operations on the rows of this matrix:
-\[ \begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix} \xrightarrow{R2\to R2-2\times R1}
+\begin{align*}
-\begin{bmatrix} 1&3&1&5\\0&1&2&7\end{bmatrix} \xrightarrow{R1\to R1-3\times R1}
+&\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}
-\begin{bmatrix} 1&0&-5&-16\\0&1&2&7\end{bmatrix}\]
+\\[6pt]\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*}
 Now we translate this back into equations to solve:
 $$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$$
@@ Line 64: / Line 68: @@
 {{page>elementary row operation}}
+==== 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{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
+\def\ar#1{\\[6pt]\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*}
+So
+  * 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 $x=9$
+The conclusion is that
+\[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}\]
+is the only solution.