Example 1 

The general equation of a straight line can be given by  

If we know two points, then we will be able to solve for a and b the unknowns. 

Here we are just interested in setting up the system of equation to find the unknowns a and b not in finding the unknowns.    

We want to find the straight line going through the points (1,2) and (-3,5).  Plugging the unknowns into the equation we arrive at the equations. 

Let us place the unknowns on the left hand side of the equation and the known on the right. 

                   or          <[1,1],[a,b]> = 2  

<[-3,1],[a,b]> = 5 

Another way of looking at the equation is the following. 

Let us look at the first equation  (recall that < *, *> is the inner product of two vectors and write the inner product as the product of a row vector times a column vector. 

<[1,1],[a,b]> =  

and  now the second 

<[-3,1],[a,b]> =
Writing the above in terms of vectors and using  vector equality we have
=
 

We now have a convention that will make things look better.  The above can be shorten to read. 

The above notation is motivated by factoring out the common term.   We have a 2 by 2 square array times a column vector equal to a column vector.   In the next section we will find that we have just observed what is ment by a new multiplication.  
To write a row vector in the with(LinearAlgebra) package use the following notation <a|b>.  The | symbol separates the columns.  If you use <a,b> you will get a column vector not a row vector.   

>	with(linalg):

 

>	with(LinearAlgebra):

 

>	r1:=<1|1>;

 

>	r2:=<-3|1>;

 

r1 and r2 represent the first and second row of  the array  and we will look at the results of multiplying the row vectors times the unknowns .  The array will be call a 2 by 2 matrix. Also the unknowns can be thought of a a matrix as well as a column vector.   

The first set of commands that follows  the vector x below be the dot probuct of r1 and x.  Using the DotProduct command from Maple will place a bar over the a and b.  You should just forget about the bar and read the dot product without the bar.  The best was to look at the dot product is by using . between the row vecto and the column vector.  This is really a multiplication of a row vector  times a column vector.  You have a row vecto with one row and two columns multiplied by a columns vector with two row and one column which will give you a quantity with one row and one column which is a scalar just like the dot product.   

>

x:=<a,b>;

 

>	DotProduct(r1,x);

 

>

r1.x;

 

(1.1.1)

 

>	DotProduct(r2,x);

 

>

r2.x;

 

(1.1.2)

 

We now use the row vectors r1 and r2 to build the matrix from the text discussion above.   

>	A:=<r1,r2>;

 

We can write the matrix A with out using the row vectors.  There are two way of building the matrix.  The first one below will list the first row and then the second row.  The second way wil be by columns.  Notice the differences.   

>	A1:=<<1|1>,<-3|1>>;

 

(1.1.3)

 

>	A2:=<<1,-3>|<1,1>>;

 

(1.1.4)

 

We finish this example by listing the system of equation as two equal vectors.  This is equivlent to the system of equations since two vectors are equal if and only if the entries in the same position are equal.   

>	A.x=<2,5>;

 

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>

 

Example 2  Quadratic equations 

Next, we  want to  find the quadratic equation that goes through three points.  

For this example we will develop a system of equations that go through the three points  

(1,3), (-2,4) and (-3, 6) 

The general equation of the quadratic is given by  

The first equation is obtained by setting x=1 and y=3 in the above quadratic.  The second equation is obtained by setting x=-2 and y=4 and the final equation is obtained by letting x=-3 and y=6. 

Now writing our system in terms of a matrices as we did in example 1 

The quantity  

is called a 3x3 matrix   

While the quantities  and are referred to as 3x1 matrices. 

We will be considering how to find the unknowns a, b and c in another lecture. The Matrix C will be built here by row.  In the next example we will use the columns to build the coefficient matrix in Maple 

>	C:=<<1|1|1>,<4|-2|1>,<9|-3|1>>;

 

>	y:=<a,b,c>;

 

>	d:=<3,4,6>;

 

>

C.y=d;

 

>

 

Example 3 

The temperature distribution in a thin wire is our next example.   

The temperature at equally spaced points along a wire can be approximated by setting the temperature at a point is equal to the average of the points on either side.  Click on the lecture2f for a development of the system of equations.  Look at fig 1 

Now using the notation we have developed in example 1 and example 2 we arrive at the following matrix system 

Notice the nice pattern of the 4x4 matrix called the coefficient matrix.  You really can not go wrong with in the way you build this matrix in Maple since the rows and columns are the same.  We will learn shortly that CT is a symmetric matrix.   

>	CT:=<<2,-1,0,0>|<-1,2,-1,0>|<0,-1,2,-1>|<0,0,-1,2>>;

 

>	TT:=<T1,T2,T3,T4>;

 

>	B:=<10,0,0,44.5>;

 

>

CT.TT=B;

 

As we know two vectors are equal if the elements are equal which is the same as looking at a system of equation. 

2T1 -T2 =10 

-T1 + 2T2 -T3 =0 

-T2 + 3T3 -T4 =0 

-T3 +2T4 = 44.5