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--- lecture_4 [2015/01/29 10:43] – created rupert
+++ lecture_4 [2015/01/29 15:08] (current) – [Discussion] rupert
@@ Line 1: / Line 1: @@
-=== Example ===
+==== 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.
+=== 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{\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}
+\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*}
+So
+  * from the last row, we get $z=-3$
+  * from the second row, we get $y=-1$
+  * from the first row, we get $x=9$
+The conclusion is again that
+\[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}\]
+is the only solution.
+=== Discussion ===
+In both of these solutions we used EROs to transform the [[augmented matrix]] into a nice form.
+  * In solution 1, we ended up with the matrix $\left[\begin{smallmatrix}1&0&3&0\\0&1&-2&5\\0&0&1&-3\end{smallmatrix}\right]$ which has a staircase pattern, with zeros below the staircase, and 1s just above the "steps" of the staircase. This is an example of a matrix in **row echelon form** (see below). We needed a bit of easy algebra, called **back substitution**, to finish off the solution. (Why is it called //echelon// form? It seems that this word has an [[http://dictionary.reference.com/browse/echelon|archaic meaning]] which is relevant to the staircase-like pattern: "any structure or group of structures arranged in a steplike form.")
+  * In solution 2, we ended up with the matrix $\left[\begin{smallmatrix}1&0&0&9\\0&1&0&-1\\0&0&1&-3\end{smallmatrix}\right]$ which has a staircase pattern with zeros below the staircase and 1s just above the "steps" of the staircase, and the additional property that we only have zeros above the 1s on the steps. This is an example of a matrix in **reduced row echelon form** (see below). Finding the solution from this matrix needed no extra algebra.
+===== Row echelon form and reduced row echelon form =====
+==== Row echelon form (REF) ====
+=== Definition ===
+{{page>zero row}}
+=== Definition ===
+{{page>leading entry}}
+=== Definition ===
+{{page>row echelon form}}
+==== Reduced row echelon form (RREF) ====
+=== Definition ===
+{{page>reduced row echelon form}}
+==== Example ====
+Use [[EROs]] to put the following matrix into [[RREF]]: \[\begin{bmatrix} 1&2&3&4&5\\0&1&2&3&4\\0&0&1&2&3\end{bmatrix}\]
+and solve the corresponding linear system.
+=== Solution ===
+\begin{align*}
+\def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}
+\def\ar#1{\\[6pt]\xrightarrow{#1}&}
+&\go{1&2&3&4&5}{0&1&2&3&4}{0&0&1&2&3}
+\ar{R2\to R2-2R3}\go{1&2&3&4&5}{0&1&0&-1&-2}{0&0&1&2&3}
+\ar{R1\to R1-3R3}\go{1&2&0&-2&-4}{0&1&0&-1&-2}{0&0&1&2&3}
+\ar{R1\to R1-2R2}\go{1&0&0&0&0}{0&1&0&-1&-2}{0&0&1&2&3}
+\end{align*}
+This matrix is in RREF. Write $x_i$ for the variable corresponding to the $i$th column. The solution is
+  - $x_4=t$, a free parameter, i.e. $t\in\mathbb{R}$. This is because the $4$th column does not contain a leading entry.
+  - From row 3: $x_3+2t=3$, so $x_3=3-2t$
+  - From row 2: $x_2-t=-2$, so $x_2=-2+t$
+  - From row 1: $x_1=0$
+So the solution is
+\[ \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\-2\\3\\0\end{bmatrix}+
+t\begin{bmatrix}0\\1\\-2\\1\end{bmatrix},\quad t\in\mathbb{R}.\]
+(Geometrically, this is a line in 4-dimensional space $\mathbb{R}^4$).