Having seen that the number $1$ plays a special role in multiplication, $1x=x$, the inverse of a number is defined as a number that multiplies by that number to give $1$. The matrix $B$ is the inverse of the matrix $A$ if when multiplied together, $A\cdot B$ or $B\cdot A$ gives the identity matrix.

Definition of Inverse Matrix

The inverse of matrix $[A]$, designated as $[A]^{-1}$, is defined by the property:

 $\displaystyle [A][A]^{-1}=[A]^{-1}[A]=[I]$

Where $[I]$ is the identity 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. The definition of an inverse matrix is based on the identity matrix $[I]$, and it has already been established that only square matrices have an associated identity matrix.

The method for finding an inverse matrix comes directly from the definition, along with a little algebra.

Finding an Inverse Matrix

Example: Find the inverse of: $\begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix}$

First, let the following be true:

$\displaystyle \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

This is the key step. It establishes $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ as the inverse that is being looked for, by asserting that it fills the definition of an inverse matrix. When multiplying this mystery matrix by our original matrix, the result is $[I]$. When solving for the four variables $a$, $b$, $c$, and $d$, then the inverse of the matrix will be found.

Next, do the multiplication:

$\displaystyle \begin{pmatrix} 3a+4c & 3b+4d \\ 5a+6c & 5b+6d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

For two matrices to be equal, every element in the left must equal its corresponding element on the right. So, for these two matrices to equal each other:

$\displaystyle 3a+4c=1$

$\displaystyle 3b+4d=0$

$\displaystyle 5a+6c=0$

$\displaystyle 5b+6d=1$

Solve the first two equations for $a$ and $c$ and the second two equations for $b$ and $d$ by using either elimination or substitution.

The results are:

$\displaystyle a=-3$

$\displaystyle b=2$

$\displaystyle c=2 \frac12$

$\displaystyle d= -1 \frac 12$

Having solved for the four variables, the result is the inverse $\begin{pmatrix} -3 & 2 \\ 2 \frac 12 & -1 \frac 12 \end{pmatrix}$.

If an inverse has been found, then a quick check to be sure it is correct is to multiply it by the original matrix and see if the identify matrix results:

$\displaystyle \begin{pmatrix} -3 & 2 \\ 2 \frac 12 & -1 \frac 12 \end{pmatrix} \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} $

After multiplying, the result is: 

$\displaystyle \begin{pmatrix} (-3)(3)+(2)(5) & (-3)(4)+(2)(6) \\ (2 \frac 12)(3)_+(-1 \frac 12)(5) & (2 \frac 12)(4)+(-1 \frac 12)(6) \end{pmatrix} $

Simplifying the problem gives:

 $\displaystyle \begin{pmatrix} -9+10 & -12+12 \\ 7 \frac 12 -7 \frac 12 & 10-9 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Therefore, the inverse works. But wait!

Note that, to fully test it, one has to try the multiplication in both orders, because, in general, changing the order of a matrix multiplication changes the answer. The definition of an inverse matrix specifies that it must work both ways. 

$\displaystyle \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix} \begin{pmatrix} -3 & 2 \\ 2 \frac 12 & -1 \frac 12 \end{pmatrix} $

After multiplying, one gets:

$\displaystyle \begin{pmatrix} (3)(-3)+(4)(2\frac 12) & (3)(2)+(4)(-1 \frac 12) \\ (5)(-3)_+(6)(2\frac 12) & (5)(2)+(6)(-1\frac 12) \end{pmatrix}$

Simplifying the problem gives:

$\displaystyle \begin{pmatrix} -9+10 & 6-6 \\ -15+15 & 10-9 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

So the inverse matrix works both ways.

In some cases, the inverse of a square matrix does not exist. This is called a singular matrix.