Multiplication of Matrices

Before we give the formal definition of how to multiply two matrices, we will discuss an example from a real life situation. Consider a city with two kinds of population: the inner city population and the suburb population. We assume that every year 40% of the inner city population moves to the suburbs, while 30% of the suburb population moves to the inner part of the city. Let I (resp. S) be the initial population of the inner city (resp. the suburban area). So after one year, the population of the inner part is

0.6 I + 0.3 S

while the population of the suburbs is

0.4 I + 0.7 S

After two years, the population of the inner city is

0.6 (0.6 I + 0.3 S) + 0.3 (0.4 I + 0.7 S)

and the suburban population is given by

0.4 (0.6 I + 0.3 S) + 0.7(0.4 I + 0.7 S)

Is there a nice way of representing the two populations after a certain number of years? Let us show how matrices may be helpful to answer this question. Let us represent the two populations in one table (meaning a column object with two entries):

\begin{displaymath}\left(\begin{array}{c}
I\\
S\\
\end{array}\right)\end{displaymath}

So after one year the table which gives the two populations is

\begin{displaymath}\left(\begin{array}{c}
0.6 I + 0.3 S\\
0.4 I + 0.7 S\\
\end{array}\right)\end{displaymath}

If we consider the following rule (the product of two matrices)

\begin{displaymath}\left(\begin{array}{cc}
a&b\\
c&d\\
\end{array}\right) \lef...
...left(\begin{array}{c}
aI + bS\\
cI + dS\\
\end{array}\right),\end{displaymath}

then the populations after one year are given by the formula

\begin{displaymath}\left(\begin{array}{cc}
0.6&0.3\\
0.4&0.7\\
\end{array}\right) \left(\begin{array}{c}
I\\
S\\
\end{array}\right).\end{displaymath}

After two years the populations are

\begin{displaymath}\left(\begin{array}{cc}
0.6&0.3\\
0.4&0.7\\
\end{array}\rig...
...ght) \left(\begin{array}{c}
I\\
S\\
\end{array}\right)\Bigg).\end{displaymath}

Combining this formula with the above result, we get

\begin{displaymath}\left(\begin{array}{cc}
0.6&0.3\\
0.4&0.7\\
\end{array}\rig...
...imes 0.4&0.4 \times 0.3 + 0.7 \times0.7\\
\end{array}\right). \end{displaymath}

In other words, we have

\begin{displaymath}\left(\begin{array}{cc}
a&b\\
c&d\\
\end{array}\right) \lef...
...{array}{cc}
ae+ bg&af+bh\\
ce + dg&cf+dh\\
\end{array}\right)\end{displaymath}

In fact, we do not need to have two matrices of the same size to multiply them. Above, we did multiply a (2x2) matrix with a (2x1) matrix (which gave a (2x1) matrix). In fact, the general rule says that in order to perform the multiplication AB, where A is a (mxn) matrix and B a (kxl) matrix, then we must have n=k. The result will be a (mxl) matrix. For example, we have

\begin{displaymath}\left(\begin{array}{ccc}
a&b&c\\
d&e&f\\
\end{array}\right)...
...a +b\beta +c\nu\\
d\alpha +e\beta +f\nu\\
\end{array}\right).\end{displaymath}

Remember that though we were able to perform the above multiplication, it is not possible to perform the multiplication

\begin{displaymath}\left(\begin{array}{c}
\alpha\\
\beta\\
\nu\\
\end{array}\...
...)\left(\begin{array}{ccc}
a&b&c\\
d&e&f\\
\end{array}\right).\end{displaymath}

So we have to be very careful about multiplying matrices. Sentences like "multiply the two matrices A and B" do not make sense. You must know which of the two matrices will be to the right (of your multiplication) and which one will be to the left; in other words, we have to know whether we are asked to perform $A \times B$ or $B \times A$. Even if both multiplications do make sense (as in the case of square matrices with the same size), we still have to be very careful. Indeed, consider the two matrices

\begin{displaymath}\left(\begin{array}{cc}
0&1\\
0&0\\
\end{array}\right)\;\mb...
...nd}\; \left(\begin{array}{cc}
0&0\\
1&0\\
\end{array}\right).\end{displaymath}

We have

\begin{displaymath}\left(\begin{array}{cc}
0&1\\
0&0\\
\end{array}\right)\left...
...ht) = \left(\begin{array}{cc}
1&0\\
0&0\\
\end{array}\right) \end{displaymath}

and

\begin{displaymath}\left(\begin{array}{cc}
0&0\\
1&0\\
\end{array}\right)\left...
...ht) = \left(\begin{array}{cc}
0&0\\
0&1\\
\end{array}\right).\end{displaymath}

So what is the conclusion behind this example? The matrix multiplication is not commutative, the order in which matrices are multiplied is important. In fact, this little setback is a major problem in playing around with matrices. This is something that you must always be careful with. Let us show you another setback. We have

\begin{displaymath}\left(\begin{array}{cc}
0&1\\
0&0\\
\end{array}\right)\left...
...egin{array}{cc}
0&0\\
0&0\\
\end{array}\right);\;\mbox{i.e.},\end{displaymath}

the product of two non-zero matrices may be equal to the zero-matrix.

More on the multiplication of matrices, may be found in the next page.

[Geometry] [Algebra] [Trigonometry ]
[Calculus] [Differential Equations] [Matrix Algebra]

S.O.S MATH: Home Page

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Author: M.A. Khamsi

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