Introduction and Basic Operations

Matrices, though they may appear weird objects at first, are a very important tool in expressing and discussing problems which arise from real life cases.

Our first example deals with economics. Indeed, consider two families A and B (though we may easily take more than two). Every month, the two families have expenses such as: utilities, health, entertainment, food, etc... Let us restrict ourselves to: food, utilities, and health. How would one represent the data collected? Many ways are available but one of them has an advantage of combining the data so that it is easy to manipulate them. Indeed, we will write the data as follows:

\begin{displaymath}\mbox{Month}=\left(\begin{array}{cccc}
\mbox{Family}&\mbox{Fo...
...ies}&\mbox{Health} \\
A&a&b&c\\
B&d&e&f\\
\end{array}\right)\end{displaymath}

If we have no problem confusing the names and what the expenses are, then we may write

\begin{displaymath}\mbox{Month}=\left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
\end{array}\right)\end{displaymath}

This is what we call a Matrix. The size of the matrix, as a block, is defined by the number of Rows and the number of Columns. In this case, the above matrix has 2 rows and 3 columns. You may easily come up with a matrix which has m rows and n columns. In this case, we say that the matrix is a (mxn) matrix (pronounce m-by-n matrix). Keep in mind that the first entry (meaning m) is the number of rows while the second entry (n) is the number of columns. Our above matrix is a (2x3) matrix.

When the numbers of rows and columns are equal, we call the matrix a square matrix. A square matrix of order n, is a (nxn) matrix.

Back to our example, let us assume, for example, that the matrices for the months of January, February, and March are

\begin{displaymath}J=\left(\begin{array}{cccc}
600&250&350\\
550&180&400\\
\en...
...0\\
625&350&410\\
\end{array}\right)\;,\;\mbox{respectively}.\end{displaymath}

To make sure that the reader knows what these numbers mean, you should be able to give the Health-expenses for family A and Food-expenses for family B during the month of February. The answers are 250 and 600. The next question may sound easy to answer, but requires a new concept in the matrix context. Indeed, what is the matrix-expense for the two families for the first quarter? The idea is to add the three matrices above. It is easy to determine the total expenses for each family and each item, then the answer is

\begin{displaymath}\mbox{First Quarter}=\left(\begin{array}{cccc}
1830&850&950\\
1775&800&1210\\
\end{array}\right)\end{displaymath}

So how do we add matrices? An approach is given by the above example. The answer is to add entries one by one. For example, we have

\begin{displaymath}\left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
\end{array}\right...
...c+ \gamma\\
d + \theta&e + \nu&f + \mu \\
\end{array}\right).\end{displaymath}

Clearly, if you want to double a matrix, it is enough to add the matrix to itself. So we have

\begin{displaymath}\mbox{double of} \left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
...
...;\left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
\end{array}\right)\end{displaymath}

which implies

\begin{displaymath}\left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
\end{array}\right...
...\begin{array}{cccc}
2a&2b&2c\\
2d&2e&2f\\
\end{array}\right).\end{displaymath}

This suggests the following rule

\begin{displaymath}2 \left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
\end{array}\rig...
...(\begin{array}{cccc}
2a&2b&2c\\
2d&2e&2f\\
\end{array}\right)\end{displaymath}

and for any number $\lambda$, we will have

\begin{displaymath}\lambda \left(\begin{array}{cccc}
a&b&c\\
d&e&f\\
\end{arra...
...lambda c\\
\lambda d&\lambda e&\lambda f\\
\end{array}\right)\end{displaymath}

Let us summarize these two rules about matrices.

Addition of Matrices: In order to add two matrices, we add the entries one by one.
Note: Matrices involved in the addition operation must have the same size.
Multiplication of a Matrix by a Number: In order to multiply a matrix by a number, you multiply every entry by the given number.

Keep in mind that we always write numbers to the left and matrices to the right (in the case of multiplication).

What about subtracting two matrices? It is easy, since subtraction is a combination of the two above rules. Indeed, if M and N are two matrices, then we will write

M-N = M + (-1)N

So first, you multiply the matrix N by -1, and then add the result to the matrix M.

Example. Consider the three matrices J, F, and M from above. Evaluate

\begin{displaymath}J + 2 F\;, \; J-M\;, \; \mbox{and} \; J-F+2M.\end{displaymath}

Answer. We have

\begin{displaymath}J + 2 F = \left(\begin{array}{cccc}
600&250&350\\
550&180&40...
...{array}{cccc}
650&330&250\\
600&270&400\\
\end{array}\right) \end{displaymath}

and since

\begin{displaymath}2 \left(\begin{array}{cccc}
650&330&250\\
600&270&400\\
\en...
...rray}{cccc}
1300&660&500\\
1200&540&800\\
\end{array}\right) \end{displaymath}

we get

\begin{displaymath}J + 2 F = \left(\begin{array}{cccc}
1900&910&850\\
1750&720&1200\\
\end{array}\right).\end{displaymath}

To compute J-M, we note first that

\begin{displaymath}(-1)M = \left(\begin{array}{cccc}
-580&-270&-350\\
-625&-350&-410\\
\end{array}\right)\;.\end{displaymath}

Since J-M = J + (-1)M, we get

\begin{displaymath}J-M = \left(\begin{array}{cccc}
600&250&350\\
550&180&400\\ ...
...gin{array}{rrrr}
20&-20&0\\
-75&-170&-10\\
\end{array}\right)\end{displaymath}

And finally, for J-F+2M, we have a choice. Here we would like to emphasize the fact that addition of matrices may involve more than one matrix. In this case, you may perform the calculations in any order. This is called associativity of the operations. So first we will take care of -F and 2M to get

\begin{displaymath}-F = (-1)F=\left(\begin{array}{cccc}
-650&-330&-250\\
-600&-...
...ay}{cccc}
1160&540&700\\
1250&700&820\\
\end{array}\right)\;.\end{displaymath}

Since J-F+2M = J + (-1)F + 2M, we get

\begin{displaymath}J-F+2M = \left(\begin{array}{cccc}
600&250&350\\
550&180&400...
...rray}{cccc}
1160&540&700\\
1250&700&820\\
\end{array}\right).\end{displaymath}

So first we will evaluate J-F to get

\begin{displaymath}J-F = \left(\begin{array}{cccc}
-50&-80&100\\
-50&-90&0\\
\end{array}\right)\end{displaymath}

to which we add 2M, to finally obtain

\begin{displaymath}J-F + 2M = \left(\begin{array}{cccc}
-50&-80&100\\
-50&-90&0...
...rray}{cccc}
1110&460&800\\
1200&610&820\\
\end{array}\right).\end{displaymath}

For the addition of matrices, one special matrix plays a role similar to the number zero. Indeed, if we consider the matrix with all its entries equal to 0, then it is easy to check that this matrix has behavior similar to the number zero. For example, we have

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

and

\begin{displaymath}\lambda \left(\begin{array}{ccc}
0&0&0\\
0&0&0\\
\end{array...
... \left(\begin{array}{ccc}
0&0&0\\
0&0&0\\
\end{array}\right).\end{displaymath}

What about multiplying two matrices? Such operation exists but the calculations involved are complicated. On the next page, we will discuss matrix multiplication.

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Author: M.A. Khamsi

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