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--- lecture_9_slides [2017/02/20 19:08] – [$A(B+C)=AB+AC$ continued] rupert
+++ lecture_9_slides [2017/02/21 10:03] (current) – rupert
@@ Line 90: / Line 90: @@
   * This is very similar, and is left as an exercise.■
-===== Matrix equations =====
-  * A linear equation can be written using [[row-column multiplication]].
-  * e.g. $ 2x-3y+z=8$ is same as $ \m{2&-3&1}\m{x\\y\\z}=8$
-  * or $ a\vec x=8$ where $a=\m{2&-3&1}$ and $\vec x=\m{x\\y\\z}$.
-==== ====
-  * We can write a whole [[system of linear equations]] in a similar way, as a matrix equation using [[matrix multiplication]].
-  * e.g. the linear system $\begin{align*} 2x-3y+z&=8\\ y-z&=4\\x+y+z&=0\end{align*}$
-  * is same as $\m{2&-3&1\\0&1&-1\\1&1&1}\m{x\\y\\z}=\m{8\\4\\0}$
-  * or $ A\vec x=\vec b$ where $A=\m{2&-3&1\\0&1&-1\\1&1&1}$, $\vec x=\m{x\\y\\z}$ and $\vec b=\m{8\\4\\0}$.
-==== ====
-In a similar way, any linear system \begin{align*} a_{11}x_1+a_{12}x_2+\dots+a_{1m}x_m&=b_1\\ a_{21}x_1+a_{22}x_2+\dots+a_{2m}x_m&=b_2\\ \hphantom{a_{11}}\vdots \hphantom{x_1+a_{22}}\vdots\hphantom{x_2+\dots+{}a_{nn}} \vdots\ & \hphantom{{}={}\!} \vdots\\ a_{n1}x_1+a_{n2}x_2+\dots+a_{nm}x_m&=b_n \end{align*}
-can be written in the form
-\[ A\vec x=\vec b\]
-where $A$ is the $n\times m $ matrix, called the **coefficient matrix** of the linear system, whose $(i,j)$ entry is $a_{ij}$ (the number in front of $x_j$ in the $i$th equation of the system).
-==== Solutions of matrix equations ====
-  * More generally, might want to solve a matrix equation like \[AX=B\] where $A$, $X$ and $B$ are matrices of any size, with $A$ and $B$ fixed matrices and $X$ a matrix of unknown variables.
-  * If $A$ is $n\times m$, we need $B$ to be $n\times k$ for some $k$, and then $X$ must be $m\times k$
-    * so we know the size of any solution $X$.
-  *  But which $m\times k$ matrices $X$ are solutions?
-==== Example ====
-If $A=\m{1&0\\0&0}$ and $B=0_{2\times 3}$, then any solution $X$ to $AX=B$ must be $2\times 3$.
-  * One solution is $X=0_{2\times 3}$
-    *  because then we have $AX=A0_{2\times 3}=0_{2\times 3}$.
-  * This is not the only solution!
-  * For example, $X=\m{0&0&0\\1&2&3}$ is another solution
-    * because then we have $AX=\m{1&0\\0&0}\m{0&0&0\\1&2&3}=\m{0&0&0\\0&0&0}=0_{2\times 3}.$
-  * So a matrix equation can have more than one solution.
-==== Example ====
-  * Let $A=\m{2&4\\0&1}$
-  * and $B=\m{3&4\\5&6}$
-  * Solve $AX=B$ for $X$
-  * $X$ must be $2\times 2$
-  * $X=\m{x_{11}&x_{12}\\x_{21}&x_{22}}$
-  * Do some algebra to solve for $X$
-  * ...
-  * Quicker way? Next time.