If $M=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$ is a 2by2 matrix with $adbc\ne 0$ then the matrix ${M}^{1}=\frac{1}{adbc}\left(\begin{array}{cc}\hfill d\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill a\hfill \end{array}\right)$ satisfies ${M}^{1}M=M{M}^{1}=I$ . We say that ${M}^{1}$ is the inverse of $M$ .
20. Inverses
20. Inverses
In the next few videos, we're going to answer the question: can you divide by a matrix?
This is the analogue of the reciprocal ${x}^{1}=1/x$ of a number: the equation ${M}^{1}M=I$ is the analogue of ${x}^{1}x=1$ .
Let's just check ${M}^{1}M=I$ (check the other equality for yourself). $$\frac{1}{adbc}\left(\begin{array}{cc}\hfill d\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)=\frac{1}{adbc}\left(\begin{array}{cc}\hfill dabc\hfill & \hfill dbbd\hfill \\ \hfill ca+ac\hfill & \hfill cb+ad\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=I.$$
We can use this to "divide" by a matrix: if we have a matrix equation $AB=C$ then we can multiply both sides on the left by ${A}^{1}$ to get ${A}^{1}AB={A}^{1}C$ , and since ${A}^{1}A=I$ this means $B={A}^{1}C$ .
With great power comes great responsibility: you should never write $\frac{B}{A}$ for matrices $A,B$ ! It's not clear if you're doing ${A}^{1}B$ or $B{A}^{1}$ (and these are different because ${A}^{1}$ and $B$ might not commute).
We can use inverses to solve simultaneous equations. For example: $$xy=1,x+y=3$$ is equivalent to $Av=b$ where $A=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)$ and $b=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \end{array}\right)$ . We can compute ${A}^{1}=\frac{1}{2}\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)$ , so $$v={A}^{1}b=\frac{1}{2}\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right),$$ and this coincides with the solution $x=1$ , $y=2$ we found earlier.
Bigger matrices
We'd like to generalise the notion of inverse to bigger matrices.
Let $A$ be an $n$ by$n$ matrix. We say that $A$ is invertible if there exists a matrix $B$ such that $AB=BA=I$ . (Here $I$ is the $n$ by$n$ identity matrix). If such a $B$ exists, then it's unique, so we're justified in calling it the inverse of $A$ and writing it as ${A}^{1}$ .
To see that the inverse is unique when it exists, suppose you had two inverses $B,C$ for $A$ : $$AB=AC=I,BA=CA=I.$$ Now evaluate $BAC$ in two ways:

$BAC=(BA)C=IC=C,$

$BAC=B(AC)=BI=B,$
so $C=BAC=B$ .
If $A,B$ are invertible matrices then $AB$ is invertible with inverse ${(AB)}^{1}={B}^{1}{A}^{1}$ .
we have $$(AB)({B}^{1}{A}^{1})=A(B{B}^{1}){A}^{1}=AI{A}^{1}=A{A}^{1}=I$$ and $$({B}^{1}{A}^{1})(AB)={B}^{1}({A}^{1}A)B={B}^{1}IB={B}^{1}B=I,$$ so ${B}^{1}{A}^{1}$ is an inverse for $AB$ .
Had we tried to use ${A}^{1}{B}^{1}$ instead, we would have obtained $AB{A}^{1}{B}^{1}$ , and we couldn't have cancelled anything because the various terms don't commute.
We saw for 2by2 matrices that a matrix is invertible if $adbc\ne 0$ . Even more basically, a 1by1 matrix $(x)$ is invertible if and only if $x\ne 0$ . We'll see in a few videos' time that there is an analogous quantity (the determinant) $det(A)$ which is nonzero precisely when $A$ is invertible. First though, we're going to explain how to calculate inverses of $n$ by$n$ matrices.