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--- lecture_6_slides [2016/02/09 12:23] – rupert
+++ lecture_6_slides [2017/02/08 17:34] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
+===== Solving linear systems. Examples; how many solutions? =====
+==== More examples ====
 ==== Example 1 ====
+If $ f(x)=ax^2+bx+c$ and $f(1)=3$, $f(2)=2$ and $f(3)=4$, find $f(x)$.
+  * $f(1)=3\implies a+b+c=3$
+  * $f(2)=2\implies 4a+2b+c=2$
+  * $f(3)=4\implies 9a+3b+c=4$
+  * $\begin{gather*}  a+b+c=3\\4a+2b+c=2\\9a+3b+c=4\end{gather*}$
+  * Solve using RREF.
+==== ====
+\begin{align*}\def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}
+\def\ar#1{\\\xrightarrow{#1}&}
+\go{1&1&1&3}{4&2&1&2}{9&3&1&4}
+\xrightarrow{R2\to R2-4R1\text{ and }R3\to R3-9R1}&
+\go{1&1&1&3}{0&-2&-3&-10}{0&-6&-8&-23}
+\ar{R2\to -\tfrac12 R2}
+\go{1&1&1&3}{0&1&\tfrac32&5}{0&-6&-8&-23}
+\ar{R3\to R3+6R2}
+\go{1&1&1&3}{0&1&\tfrac32&5}{0&0&1&7}
+\end{align*}
+  * So far: in REF!
+==== ====
+\begin{align*}
+\go{1&1&1&3}{0&1&\tfrac32&5}{0&0&1&7}
+\xrightarrow{R1\to R1-R3\text{ and }R2\to R2-\tfrac32R3}&
+\go{1&1&0&-4}{0&1&0&-5.5}{0&0&1&7}
+\ar{R1\to R1-R2}
+\go{1&0&0&1.5}{0&1&0&-5.5}{0&0&1&7}
+\end{align*}
+  * So $a=1.5$, $b=-5.5$ and $c=7$
+  * So $f(x)=1.5x^2-5.5x+7$.
+==== Example 2 ====
 Solve the linear system
@@ Line 36: / Line 73: @@
   * No free variables.
-==== Example 2 ====
+==== Example 3 ====
 Solve the linear system
@@ Line 64: / Line 101: @@
   * To detect this: put in REF and find a row $[0~0~\dots~0~1]$.
-==== Example 3 ====
+==== Example 4 ====
 For which value(s) of $k$ does the following linear system have
@@ Line 114: / Line 151: @@
     * the $(1,1)$ entry of $B$ is $b_{11}=99$
     * the $(1,3)$ entry of $B$ is $b_{13}=5$
-    * the $(2,1)$ entry is $b_{21}=7$, etc.
+    * the $(2,1)$ entry of $B$ is $b_{21}=7$
+    * etc.
+  * $(3,2)$ entry of $B$?
+    * undefined!
 ==== =====
   * $\left[\begin{smallmatrix}3\\2\\4\\0\\-1\end{smallmatrix}\right]$ is a $5\times 1$ matrix.
@@ Line 122: / Line 162: @@
   * Even though these have the same entries, they have a different "shape", or "size" and they are different matrices.
-=== Size of a matrix ===
+==== Size of a matrix ====
 {{page>same size}}
-=== Equality of matrices ===
+==== Equality of matrices ====
 {{page>equal matrices}}
@@ Line 133: / Line 173: @@
   * $\begin{bmatrix}3\\2\\4\\0\\-1\end{bmatrix}\ne \begin{bmatrix}3&2&4&0&-1\end{bmatrix}$, since these matrices have different sizes: the first is $5\times 1$ but the second is $1\times 5$.
-  * $\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}1 &0\\2&0\end{bmatrix}$ since these matrices are not the same size.
+==== ====
-  * $\begin{bmatrix}1&0\\0&1\end{bmatrix}\ne \begin{bmatrix}1&0\\1&0\end{bmatrix}$ because even though they have the same size, the $(2,1)$ entries are different.
+  * $\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}1 &0\\2&0\end{bmatrix}$
-  * If $\begin{bmatrix}3x&7y+2\\8z-3&w^2\end{bmatrix}=\begin{bmatrix}1&2z\\\sqrt2&9\end{bmatrix}$ then we know that all the corresponding entries are equal, so we get four equations:\begin{align*}3x&=1\\7y+2&=2z\\8z-3&=\sqrt2\\w^2&=9\end{align*}
+    * not the same size.
+  * $\begin{bmatrix}1&0\\0&1\end{bmatrix}\ne \begin{bmatrix}1&0\\1&1\end{bmatrix}$
+    * same size but the $(2,1)$ entries are different.
+==== ====
+  * If $\begin{bmatrix}3x&7y+2\\8z-3&w^2\end{bmatrix}=\begin{bmatrix}1&2z\\\sqrt2&9\end{bmatrix}$ then we know that all the corresponding entries are equal
+  * We get four equations:\begin{align*}3x&=1\\7y+2&=2z\\8z-3&=\sqrt2\\w^2&=9\end{align*}