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--- lecture_3 [2015/01/27 11:59] – rupert
+++ lecture_3 [2016/02/02 10:05] (current) – rupert
@@ Line 1: / Line 1: @@
-Let's look at this example more closely:
+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 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 33: / Line 37: @@
 Let's give some terminology which will allow us to make this process clear.
-===== The augmented matrix of a linear system =====
+===== The augmented matrix of a system of linear equations =====
 ==== Definition ====
@@ Line 51: / Line 55: @@
 \end{align*}
 so the augmented matrix is
-\[ \begin{bmatrix} 3&4&7&2\\1&0&3&0\\0&1&-2&5\end{bmatrix}\]
+\[ \begin{bmatrix} 3&4&7&2\\1&0&3&0\\0&1&-2&5\end{bmatrix}.\]
+  * the $(2,3)$ entry of this matrix is $3$;
+  * the $(3,2)$ entry is $1$;
+  * the $(1,4)$ entry is $2$;
+  * the $(4,1)$ entry is undefined (since this matrix does not have a $4$th row).
+===== Elementary operations on a system of linear equations =====
+{{page>elementary operations on a linear system}}
+===== Elementary row operations on a matrix =====
+{{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.