10.3 The inverse matrix

We have seen that the number 1 plays a special role in multiplication, because $1x=x$ .

The inverse of a number is defined as the number that multiplies by that number to give 1: $b$ is the inverse of $a$ if $ab=1$ . Hence, the inverse of $3$ is $\frac{1}{3}$ ; the inverse of $\frac{-5}{8}$ $=$ $\frac{-8}{5}$ . Every number except 0 has an inverse.

By analogy, the inverse of a matrix multiplies by that matrix to give the identity matrix.

Definition of inverse matrix

The superscript –1 is being used here in a similar way to its use in functions. Recall that $f^{\mathrm{–1}}\left(\mathrm{x}\right)$ does not designate an exponent of any kind, but instead, an inverse function. In the same way, $[A{]}^{\mathrm{–1}}$ does not denote an exponent, but an inverse matrix .

Note that, just as in the definition of the identity matrix, this definition requires commutativity—the multiplication must work the same in either order.

Note also that only square matrices can have an inverse. Why? The definition of an inverse matrix is based on the identity matrix $\left[I\right]$ , and we already said that only square matrices even have an identity!

How do you find an inverse matrix? The method comes directly from the definition, with a little algebra.

Did it work? Let’s find out.

Testing our Inverse Matrix
$\left[\begin{array}{cc}-3& 2\\ 2\frac{1}{2}& -1\frac{1}{2}\end{array}\right]$ $\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$ $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$	The definition of an inverse matrix: if we have indeed found an inverse, then when we multiply it by the original matrix, we should get [I].
$\left[\begin{array}{cc}(-3)(3)+(2)(5)& (-3)(4)+(2)(6)\\ \left(2\frac{1}{2}\right)(3)+\left(-1\frac{1}{2}\right)(5)& \left(2\frac{1}{2}\right)(4)+\left(-1\frac{1}{2}\right)(6)\end{array}\right]$ $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$	Do the multiplication.
$\left[\begin{array}{cc}-9+\text{10}& -\text{12}+\text{12}\\ 7\frac{1}{2}-7\frac{1}{2}& \text{10}-9\end{array}\right]$ = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$	It works!

Note that, to fully test it, we would have to try the multiplication in both orders . Why? Because, in general, changing the order of a matrix multiplication changes the answer; but the definition of an inverse matrix specifies that it must work both ways! Only one order was shown above, so technically, we have only half-tested this inverse.

This process does not have to be memorized: it should make logical sense. Everything we have learned about matrices should make logical sense, except for the very arbitrary-looking definition of matrix multiplication.