It is important to always think of a matrix as a representation of the transformed standard basis vectors rather than just thinking about a matrix as a rectangular array of random real numbers.

With this thought, matrix multiplication can be visualised as applying a series of transformations from right to left. For example, Lets say,
A=\begin{bmatrix}  1 &2 \\   3 &4   \end{bmatrix},  B=\begin{bmatrix}  5 &6 \\   7 &8   \end{bmatrix}
Then the matrix product AB is equivalent to first applying the transformation B and then applying the transformation A.

But on which vectors are we applying these transformations?

Well, its always the standard basis vectors (\hat{i} and \hat{j} in this case) which are transformed (while keeping the origin constant). We can attach the matrix of standard basis vectors next to the matrix B to signify that, first the transformation B is applied on the standard basis vectors and, then the transformation A is applied on the resultant matrix.
Let S denote the 2×2 standard basis vector matrix,
S=\begin{bmatrix}  1 &0 \\    0& 1  \end{bmatrix}
Now the matrix multiplication can be written as, ABS
ABS=\begin{bmatrix}  1 &2 \\    3& 4  \end{bmatrix}  \begin{bmatrix}  5 & 6\\    7&8   \end{bmatrix}  \begin{bmatrix}  1 &0 \\    0& 1  \end{bmatrix}
Starting from the right, lets first calculate the product BS.
We will find the transformations for the unit vectors \hat{i} and \hat{j} separately as vectors and then combine them into a matrix just like S. So, the transformation for the unit vector \hat{i} is,
\begin{bmatrix}  5 &6 \\    7&8   \end{bmatrix}  \begin{bmatrix}  1\\   0  \end{bmatrix}=  1\begin{bmatrix}  5\\   7  \end{bmatrix}+  0\begin{bmatrix}  6\\   8  \end{bmatrix}=  \begin{bmatrix}  5\\   7  \end{bmatrix}
and the transformation for the unit vector \hat{j} is,
\begin{bmatrix}  5 &6 \\    7&8   \end{bmatrix}  \begin{bmatrix}  0\\   1  \end{bmatrix}=  0\begin{bmatrix}  5\\   7  \end{bmatrix}+  1\begin{bmatrix}  6\\   8  \end{bmatrix}=  \begin{bmatrix}  6\\   8  \end{bmatrix}
Combining the above results in a matrix,
BS=\begin{bmatrix}  5 &6 \\    7& 8  \end{bmatrix}
The next transformation matrix A is now applied on the product BS
\begin{bmatrix}  1 &2 \\   3& 4  \end{bmatrix}  \begin{bmatrix}  5 &6 \\   7 &8   \end{bmatrix}
Again, we will find the transformations for the unit vectors \hat{i} and \hat{j} separately as vectors and then combine them into a matrix. So, the transformation for the unit vector \hat{i} is,
\begin{bmatrix}  1 &2 \\   3& 4  \end{bmatrix}  \begin{bmatrix}  5\\   7  \end{bmatrix}=  5\begin{bmatrix}  1\\   3  \end{bmatrix}+  7\begin{bmatrix}  2\\   4  \end{bmatrix}=  \begin{bmatrix}  5\\   15  \end{bmatrix}+  \begin{bmatrix}  14\\   28  \end{bmatrix}=  \begin{bmatrix}  19\\   43  \end{bmatrix}
and the transformation for the unit vector \hat{j} is,
\begin{bmatrix}  1 &2 \\    3&4   \end{bmatrix}  \begin{bmatrix}  6\\   8  \end{bmatrix}=  6\begin{bmatrix}  1\\   3  \end{bmatrix}+  8\begin{bmatrix}  2\\   4  \end{bmatrix}=  \begin{bmatrix}  6\\   18  \end{bmatrix}+  \begin{bmatrix}  16\\   32  \end{bmatrix}=  \begin{bmatrix}  22\\   50  \end{bmatrix}
Combining the above results in a matrix,
ABS=AB = \begin{bmatrix}  19 & 22\\    43& 50  \end{bmatrix}

What is the significance of matrix multiplication

Comments, Questions or Suggestions: