The number $1$ has a special property: when multiplying any number by $1$, the result is the same number, i.e. $5 \cdot 1 = 5$.  This idea can be expressed with the following property as an algebraic generalization: $1x=x$.  The matrix that has this property is referred to as the identity matrix.

Definition of the Identity Matrix

The identity matrix, designated as $[I]$, is defined by the property: 

$\displaystyle [A][I]=[I][A]=[A]$.

Note that the definition of [I][I] stipulates that the multiplication must commute, that is, it must yield the same answer no matter in which order multiplication is done.

This stipulation is important because, for most matrices, multiplication does not commute.

What matrix has this property? A first guess might be a matrix full of $1$s, but that does not work:

$\displaystyle \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 7 & 7 \end{pmatrix}$ 

So $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ is not an identity matrix.

The matrix that does work is a diagonal stretch of $1$s, with all other elements being $0$.

$\displaystyle \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$

So $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix for $2 \times 2$ matrices.

For a $3 \times 3$ matrix, the identity matrix is a $3 \times 3$ matrix with diagonal $1$s and the rest equal to $0$:

$\displaystyle \begin{pmatrix} 2 & \pi & -3 \\ 5 & -2 & \frac 12 \\ 9 & 8 & 8.3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & \pi & -3 \\ 5 & -2 & \frac 12 \\ 9 & 8 & 8.3 \end{pmatrix}$

So $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ is the identity matrix for $3 \times 3$ matrices.

It is important to confirm those multiplications, and also confirm that they work in reverse order (as the definition requires). 

There is no identity for a non-square matrix because of the requirement of matrices being commutative. For a non-square matrix $[A] $ one might be able to find a matrix $[I]$ such that $[A][I]=[A]$, however, if the order is reversed then an illegal multiplication will be left. The reason for this is because, for two matrices to be multiplied together, the first matrix must have the same number of columns as the second has rows.