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

We've seen that a single linear equation can be written using [[row-column multiplication]]. For example,
\[ 2x-3y+z=8\]
can be written as 
\[ \def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{2&-3&1}\m{x\\y\\z}=8\]
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]]. For example we can rewrite the linear system
\begin{align*} 2x-3y+z&=8\\ y-z&=4\\x+y+z&=0\end{align*}
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 $\vec b=\m{8\\4\\0}$. (We are writing the little arrow above the column vectors here because otherwise we might get confused between the $\vec x$: a column vector of variables, and $x$: just a single variable).


More generally, 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) and $\vec x=\m{x_1\\x_2\\\vdots\\x_m}$, and $\vec b=\m{b_1\\b_2\\\vdots\\b_n}$.

More generally still, we 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. Because of the definition of [[matrix multiplication]], 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 $A=\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{1&0\\0&0}$ and $B=0_{2\times 3}$, then any solution $X$ to $AX=B$ must be $2\times 3$.

One solution is $X=0_{2\times 3}$, since in this case we have $AX=A0_{2\times 3}=0_{2\times 3}$.

However, this is not the only solution. For example, $X=\m{0&0&0\\1&2&3}$ is another solution, since in this case \[AX=\m{1&0\\0&0}\m{0&0&0\\1&2&3}=\m{0&0&0\\0&0&0}=0_{2\times 3}.\]

So from this example, we see that a matrix equation can have many solutions.

===== Invertibility =====

We've seen that solving matrix equations $AX=B$ is useful, since they generalise systems of linear equations.

How can we solve them?

==== Example ====

Take $A=\def\mat#1{\begin{bmatrix}#1\end{bmatrix}}\mat{2&4\\0&1}$ and $B=\mat{3&4\\5&6}$, so we want to find all matrices $X$ so that $AX=B$, or \[ \mat{2&4\\0&1}X=\mat{3&4\\5&6}.\] Note that $X$ must be a $2\times 2$ matrix for this to work, by the definition of [[matrix multiplication]]. So one way to solve this is to write $X=\mat{x_{11}&x_{12}\\x_{21}&x_{22}}$ and plug it in:
\[\mat{2&4\\0&1}\mat{x_{11}&x_{12}\\x_{21}&x_{22}}=\mat{3&4\\5&6}\iff \mat{2x_{11}+4x_{21}&2 x_{12}+4x_{22}\\x_{21}&x_{22}}=\mat{3&4\\5&6}\]
and then equate entries to 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*}
which we can solve in the usual way.

But this is a bit tedious! We will develop a slicker method by first thinking about solving ordinary equations $ax=b$ where $a,x,b$ are all numbers, or  if you like, $1\times 1$ matrices.

==== Solving $ax=b$ and $AX=B$ ====

If $a\ne0$, then solving $ax=b$ where $a,b,x$ are numbers is easy. We just divide both sides by $a$, or equivalently, we multiply both sides by $\tfrac1a$, to get the solution: $x=\tfrac1a\cdot b$.

Why does this work? If $x=\tfrac1a\cdot b$, then
\begin{align*} ax&=a(\tfrac1a\cdot b)\\&=(a\cdot \tfrac1a)b\\&=1b\\&=b\end{align*}
so $ax$ really is equal to $b$, and we do have a solution to $ax=b$.

What is special about $\tfrac1a$ which made this all work?

  - we have $a\cdot \tfrac1a = 1$,
  - and $1b = b$.

Now for an $n\times k$ matrix $B$, we know that the identity matrix $I_n$ does the same sort of thing as $1$ is doing in the relation $1b=b$: we have $I_nB=B$ for any $n\times k$ matrix $B$. So instead of $\tfrac1a$, we want to find a matrix $C$ with the property: $AC=I_n$. In fact, because matrix multiplication is not commutative, we also require that $CA=I_n$. It's then easy to argue that $X=C\cdot B$ is a solution to $AX=B$, since
\begin{align*} AX&=A(CB)\\&=(AC)B\\&=I_nB\\&=B.\end{align*}

==== Example revisited ====
If $A=\mat{2&4\\0&1}$, then the matrix $C=\mat{\tfrac12&-2\\0&1}$ does have the property that
\[ A C=I_2=C A.\]
(You should check this!). So a solution to $AX=B$ where $B=\mat{3&4\\5&6}$ is $X=CB=\mat{\tfrac12&-2\\0&1}\mat{3&4\\5&6} = \mat{-8.5&-10\\5&6}$.

Notice that having found the matrix $C$, then we can solve $AX=C$ easily for any $2\times 2$ matrix $C$: the answer is $X=CC$. This is quicker than having to solve four new linear equations using our more tedious method above.

==== Definition: invertible ====

{{page>invertible}}

==== Examples ====

  * $A=\mat{2&4\\0&1}$ is invertible, and the matrix $C=\mat{\tfrac12&-2\\0&1}$ is an inverse of $A$
  * 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 $n$, since $I_n\cdot I_n=I_n=I_n\cdot I_n$, so an inverse of $I_n$ is $I_n$.
  * $0_{n\times n}$ is not invertible for any $n$, since $0_{n\times n}\cdot C=0_{n\times n}$ for any $n\times n$ matrix $C$, so $0_{n\times n}\cdot C\ne I_n$.
  * $A=\mat{1&0\\0&0}$ is not invertible, since for any $2\times 2$ matrix $C=\mat{a&b\\c&d}$ we have $AC=\mat{a&b\\0&0}$ which is not equal to $I_2=\mat{1&0\\0&1}$ since the $(2,2)$ entries are not equal.
  * $A=\mat{1&2\\-3&-6}$ 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 ===
Suppose $C$ and $C'$ are both inverses of $A$. Then
$AC=I_n=CA$ and $AC'=I_n=C'A$. So
\begin{align*} C&=CI_n\quad\text{by the properties of $I_n$}\\&=C(AC')\quad\mbox{because }AC'=I_n\\&=(CA)C'\quad\mbox{because matrix multiplication is associative}\\&=I_nC'\quad\mbox{because }CA=I_n\\&=C'\quad\text{by the properties of $I_n$}.\end{align*}
So $C=C'$, whenever $C$ and $C'$ are inverses of $A$. So $A$ has a unique inverse. ■

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

{{page>the inverse}}

If a matrix $A$ is not invertible, then $A^{-1}$ does not exist.

=== Warning ===

If $A$ is a matrix then $\frac 1A$ doesn't make sense! You should never write this down. In particular, $A^{-1}$ definitely doesn't mean $\frac 1A$.

Similarly, you should **never** write down $\frac AB$ where $A$ and $B$ are matrices. This doesn't make sense either!
==== Examples revisited ====

  * $A=\mat{2&4\\0&1}$ has $A^{-1}=\mat{\tfrac12&-2\\0&1}$. In other words, $\mat{2&4\\0&1}^{-1}=\mat{\tfrac12&-2\\0&1}$.
  * a $1\times 1$ matrix $A=[a]$ with $a\ne 0$ has $[a]^{-1}=[\tfrac1a]$.
  * $I_n^{-1}=I_n$.
  * $0_{n\times n}^{-1}$ does not exist
  * $\mat{1&0\\0&0}^{-1}$ does not exist
  * $\mat{1&2\\-3&-6}^{-1}$ does not exist