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):

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

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

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

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

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

then the populations after one year are given by the formula

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

After two years the populations are

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

Combining this formula with the above result, we get

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

In other words, we have

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

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

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

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

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

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

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

We have

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

and

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

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

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

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]

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Author: M.A. Khamsi

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