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--- lecture_3 [2015/01/27 12:09] – [Elementary operations on a system of linear equations] 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 60: / Line 64: @@
 ===== Elementary operations on a system of linear equations =====
-If we perform one of the following operations on a system of linear equations:
+{{page>elementary operations on a linear system}}
+===== Elementary row operations on a matrix =====
-  - list the equations in a different order; or
+{{page>elementary row operation}}
-  - multiply one of the equations by a non-zero real number; or
-  - replace equation $j$ by equation $j$ ${}+{}$ $c\times {}$ equation $i$, where $c$ is a non-zero real number,
-then the new system will have exactly the same solutions as the original system. For operations of type 1, this is trivial, and for operations of types 2 and 3, this is because
-  * we are doing 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; and
+==== Example ====
-  * these operations are reversible, using operations of the same type, so any solution to the new system will also be a solution to the original system.
-We call these operations of types 1, 2 or 3 **elementary operations** on the linear system.
+Use [[EROs]] to find the intersection of the planes
+\begin{align*} 3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}
-===== Elementary row operations on a matrix =====
+=== Solution 1 ===
-Recall that when we form the augmented matrix of a linear system, each equation in the system becomes a row of the matrix. So we can translate the elementary operations of the linear system into corresponding operations on the rows of the matrix. We get three different types:
+\begin{align*}
+\def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
-  - change the order of the rows of the matrix;
+\def\ar#1{\\[6pt]\xrightarrow{#1}&}
-  - multiply one of the rows of the matrix by a non-zero real number;
+&\go{3&4&7&2}{1&0&3&0}{0&1&-2&5}
-  - replace row $j$ by row $j$ ${}+{}$ $c\times {}$ row $i$, where $c$ is a non-zero real number.
+\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*}
-The system of linear equations corresponding to these matrices will then have exactly the same solutions.
+So
-We call these operations **elementary row operations** on the matrix.
+  * 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.