~~REVEAL~~


===== Matrix equations =====

  * A linear equation can be written using [[row-column multiplication]]. 
  * e.g. $ \newcommand{\m}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]} 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 $X$
  * ...
  * Is there a quicker way? 

==== Example ====

  * Consider $AX=B$, where
    * $A=\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\m{1&0\\0&0}$
    * $B=0_{2\times 3}$,
  * i.e. $\m{1&0\\0&0}X=\m{0&0&0\\0&0&0}$
  * then any solution $X$ to $AX=B$ must be $2\times 3$.
  * One solution is $X=0_{2\times 3}$
  * Are there any more?
  * Yes! e.g. $X=\m{0&0&0\\1&2&3}$
  * So a matrix equation can have more than one solution.

===== Invertibility ======

==== Example ====

  * Take $A=\def\mat#1{\m{#1}}\mat{2&4\\0&1}$ and $B=\mat{3&4\\5&6}$, consider $AX=B$
  * $\mat{2&4\\0&1}X=\mat{3&4\\5&6}$
  * $X$ must be $2\times 2$ matrix
  * One way to solve: write $X=\mat{x_{11}&x_{12}\\x_{21}&x_{22}}$ 
  * Plug in and do matrix multiplication: $\mat{2x_{11}+4x_{21}&2 x_{12}+4x_{22}\\x_{21}&x_{22}}=\mat{3&4\\5&6}$
  * Get four linear equations: $\begin{align*}2x_{11}+4x_{21}&=3\\2 x_{12}+4x_{22}&=4\\x_{21}&=5\\x_{22}&=6\end{align*}$
  * Can solve in the usual way....
  * Tedious! Can we do better?

==== Simpler: $1\times 1$ matrix equations ====

  * Let $a,b$ be ordinary numbers ($1\times 1$ matrices) with $a\ne0$. How do we solve $ax=b$?
  * Answer: multiply both sides by $a^{-1}$ 
    * (for numbers, $a^{-1}$ is same as $\tfrac1a$)
  * Solution: $x=a^{-1}b$.

  * Why does this work?
  * If $x=\tfrac1a\cdot b$, then $ax=a(\tfrac1a\cdot b)=(a\cdot \tfrac1a)b=1b=b$
  * so $ax$ really is equal to $b$
  * We do have a solution to $ax=b$.

==== Thinking about $a^{-1}$ for $a$ a number ====

  * What is special about $a^{-1}$ which made this all work?
  * Write $c=a^{-1}$

  * $1b=b$, and $a c = 1$
  * so $ x=cb$ has $ax=acb=1b=b$ :-)

==== What about matrices? ====

  * Can we do something similar for an $n\times n$ matrix $A$?
  * i.e. find a matrix $C$ with similar properties?
    * Replace $1$ with $I_n$
    * Then $I_nB=B$ 
    * So if we had a matrix $C$ with $CA=I_n$...
    * Then $X=CB$ would have $AX=ACB=I_nB=B$ :-)
  * The key property of $C$ is that $AC=I_n$.
  * Turns out we also want $CA=I_n$

==== Example revisited ====
  * Let $A=\mat{2&4\\0&1}$
  * The matrix $C=\mat{\tfrac12&-2\\0&1}$ **does** have the property\[ C A =I_2= AC.\]
  * Check this!
  * Multiply both sides of $AX=B$ on the left by $C$ and use $CA=I_2$:
  * Get $X=CB=\mat{\tfrac12&-2\\0&1}\mat{3&4\\5&6} = \mat{-8.5&-10\\5&6}$.
  * This is quicker than solving lots of equations
    * although we don't yet know how to **find** $C$ from $A$
  * Also, if we change $B$ we can easily find solutions to the new equation

==== Definition: invertible ====

{{page>invertible}}

==== Examples ====

  * $A=\mat{2&4\\0&1}$ is invertible, and $C=\mat{\tfrac12&-2\\0&1}$ is an inverse
  * a $1\times 1$ matrix $A=[a]$ is invertible if and only if $a\ne0$, and if $a\ne0$ then an inverse of $A=[a]$ is $C=[\tfrac1a]$.
  * $I_n$ is invertible for any $n$, with inverse $I_n$
  * $0_{n\times n}$ is not invertible for any $n$... why?
  * $A=\mat{1&0\\0&0}$ is not invertible... why?
  * $A=\mat{1&2\\-3&-6}$ is not invertible. We'll see why later!

==== Proposition: uniqueness of the inverse ====
If $A$ is an invertible $n\times n$ matrix, then $A$ has a //unique// inverse.

=== Proof ===
  * $A$ invertible, so $A$ has at least one inverse.
  * Suppose it has two inverses, say $C$ and $D$.
  * Then $AC=I_n=CA$ and $AD=I_n=DA$.
  * So $C=CI_n=C(AD)=(CA)D=I_nD=D$
  * So $C=D$.
  * So any two inverses of $A$ are equal.
  * So $A$ has a unique inverse.■

==== Definition/notation: $A^{-1}$ ====

{{page>the inverse}}

==== Examples again ====

  * $A=\mat{2&4\\0&1}$ is invertible, and $A^{-1}=\mat{\tfrac12&-2\\0&1}$.
    * i.e. $\mat{2&4\\0&1}^{-1}=\mat{\tfrac12&-2\\0&1}$
  * if $a$ is a non-zero scalar, then $[a]^{-1}=[\tfrac 1a]$
  * $I_n^{-1}=I_n$ for any $n$
  * $0_{n\times n}^{-1}$, $\mat{1&0\\0&0}^{-1}$ and $\mat{1&2\\-3&-6}^{-1}$ **do not exist** 
    * because these matrices aren't invertible
==== Warning ====

If $A$, $B$ are matrices, then $\frac AB$ doesn't make sense! 

  * Never write this down as it will almost always lead to mistakes.
  * In particular, $A^{-1}$ definitely doesn't mean $\frac 1A$. Never write this either!

==== Proposition: solving $AX=B$ when $A$ is invertible ====

If $A$ is an invertible $n\times n$ matrix and $B$ is an $n\times k$ matrix, then $ AX=B$ has a unique solution: $X=A^{-1}B$.

=== Proof ===

  * First check that $X=A^{-1}B$ really is a solution:
    * $AX=A(A^{-1}B)=(AA^{-1})B=I_nB=B$ :-)
  * Uniqueness: suppose $X$ and $Y$ are both solutions
  * Then $AX=B$ and $AY=B$, so $AX=AY$.
  * Multiply both sides on the left by $A^{-1}$:
    * $A^{-1}AX=A^{-1}AY$, or $I_nX=I_n Y$, or $X=Y$.
  * So any two solutions are equal.
  * So $AX=B$ has a unique solution.■

==== Corollary ====

If $A$ is an $n\times n$ matrix and there is a non-zero $n\times m$ matrix $K$ so that $AK=0_{n\times m}$, then $A$ is not invertible.

=== Proof ===

  * $AX=0_{n\times m}$ has (at least) two solutions: 
    * $X=K$, 
    * and $X=0_{n\times m}$ 
  * If $A$ was invertible, this would contradict the Proposition. 
  * So $A$ cannot be invertible. ■

==== Example ====

  * Let $A=\mat{1&2\\-3&-6}$. $A$ isn't invertible... why?
  * One column of $A$ is $2$ times the other... exploit this.
  * Let $K=\mat{-2\\1}$ 
  * $K$ is non-zero but $AK=\mat{1&2\\-3&-6}\mat{-2\\1}=\mat{0\\0}=0_{2\times 1}$
  * So $A$ is not invertible, by the Corollary.

  * Next time: a more systematic way to determine when a matrix is invertible: **determinants**