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--- lecture_7a [2015/02/10 10:49] – rupert
+++ lecture_7a [2015/02/10 10:55] (current) – [One more example] rupert
@@ Line 3: / Line 3: @@
 Solve the following [[linear system]]:
 \begin{align*} x+y+z&=2\\2x-z&=0\\x+3y+4z&=6\\3x+y&=2\end{align*}
+== Solution ==
+We remark that there are more equations than unknowns... but this isn't a problem! We proceed as usual:
 \begin{align*}
 \def\go#1#2#3#4{\begin{bmatrix}#1\\#2\\#3\\#4\end{bmatrix}}
 \def\ar#1{\\[6pt]\xrightarrow{#1}&}
-&\go{3&4&-2&1}{1&1&1&4}{2&5&1&3}
+&\go{1&1&1&2}{2&0&-1&0}{1&3&4&6}{3&1&0&2}
-\ar{R1\leftrightarrow R2}
+\ar{R2\to R2-2R1,\ R3\to R3-R1\text{ and }R4\to R4-3R1}
-\go{1&1&1&4}{3&4&-2&1}{2&5&1&3}
+\go{1&1&1&2}{0&-2&-3&-4}{0&2&3&4}{0&-2&-3&-4}
-\ar{R2\to R2-3R1\text{ and }R3\to R3-2R1}
+\ar{R3\to R3+R2\text{ and }R4\to R4-R2}
-\go{1&1&1&4}{0&1&-5&-11}{0&3&-1&-5}
+\go{1&1&1&2}{0&1&1.5&2}{0&0&0&0}{0&0&0&0}
-\ar{R3\to R3-3R1}
-\go{1&1&1&4}{0&1&-5&-11}{0&0&14&28}
-\ar{R3\to \tfrac1{14}R3}
-\go{1&1&1&4}{0&1&-5&-11}{0&0&1&2}
-\ar{R1\to R1-R3\text{ and }R2\to R2+5R3}
-\go{1&1&0&2}{0&1&0&-1}{0&0&1&2}
 \ar{R1\to R1-R2}
-\go{1&0&0&3}{0&1&0&-1}{0&0&1&2}
+\go{1&0&-0.5&0}{0&1&1.5&2}{0&0&0&0}{0&0&0&0}
 \end{align*}
-== Solution ==
+Now $z$ is a free variable, say $z=t$ where $t\in \mathbb{R}$, and from row 2 we get $y=2-1.5t$ and row 1 gives $x=0.5t$, so the solution is
+\[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\2\\0\end{bmatrix}+t\begin{bmatrix}0.5\\-1.5\\1\end{bmatrix},\quad t\in\mathbb{R}.\]
-We remark that there are more equations than unknowns... but this isn't a problem! We proceed as usual:
+===== Observations about Gaussian elimination =====
 {{page>gaussian elimination remarks}}