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Systems of Linear Equations. Investigate Systems of linear equations where the system constants have well known mathematical patterns.

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Mathematical Investigation

Investigate Systems of linear equations where the system constants have well known mathematical patterns.

Part A:image00.pngimage01.png

Here is a 2x2 system of linear equations:            

It is easy to notice that if one starts from the coefficient of x, they can generate the coefficient of y by adding 1 to it and the constant by adding 1 to the coefficient of y. Another way to look at this is to start from the constant and add -1 to it you can generate the y coefficient and add -1 to this to get the x coefficient. The same is true for the second equation except that instead of adding 1 to the coefficient of x, we add -3. Similarly, we can start from the constant and add 3 each time to get the y and x coefficients.


       x + 2y= 3           x= -2y + 3                                                                      x= -2(2) + 3        x= -1image06.pngimage06.png

       2x – y= -4         2(-2y + 3) – y= -4        -4y + 6 – y= -4       -5y= -10          y= 2                    y= 2

The solution x equals -1 and y equals 2 means that that point satisfies both equations. As we can see in the graph the two equations intersect at a point and that point is the same as the solution we found when we solved the problem algebraically. So the point that the two equations intersect is (-1, 2).


Here are some examples using the same pattern:

Example 1:

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c(a+ b) – cay – cby + cay + day= a(c + 2d)     ca + 2cb – cby + day= ca + 2ad     2cb – cby + day= 2ad     image30.pngimage29.pngimage26.pngimage28.png


                                                   ax= a + 2b – (a + b)(2)      ax= a + 2b – 2a – 2bimage23.pngimage26.pngimage26.png

day – cby= 2ad – 2cb      y(ad – bc)= 2(ad – bc)        y= 2                          y=2          image34.png

      a(1 – 2)         x= -1


y=2                      y= 2

We can see that even in the most general form the answers are the same.

Originally I guessed that the 3x3 system of equations would work the same way as the 2x2 system. I would have thought that all the 3x3 systems would have a common solution. Unlike the 2x2 system of equations for which I used the method of substitution to solve, I used matrix algebra to solve the 3x3 system.

Let’s consider a 3x3 system of linear equations:

               x + 2y + 3z= 4                                         1   2   3         x              4

2x – y – 4z= -7                                           2  -1  -4        y       =      -7         or   AX=B

3x + 5y + 7z= 10                                       3   5   7         z               10

              By multiplying both sides by A-1 , the inverse matrix of A, we get the solution. In other words,

A-1 AX= A-1BIX=A-1 B           X= A-1B   (where I is the 3x3 identity matrix)

All we have to do is find the inverse matrix of A, which I did using a graphing calculator, as shown in picture 1. To my surprise I found that the inverse did not exist.

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0px;" alt="image38.png" />image38.png         y= bx - image38.pngimage38.png

As the value of a goes from - ∞       zero the slope increases from zero        + ∞, while the y-intercept is always negative and it decreases in absolute value. The x-intercept is equal to a2 and decreases as the value of a decreases. When a becomes positive, the slopes of the lines are negative. Actually the slopes increase from - ∞         zero as the a goes from zero       + ∞.  

We observe a similar pattern in the other family of equations. We notice that we get the same line for a= - image39.pngimage39.png as for b = 2; for a = - image40.pngimage40.png  as for b = 5; etc. IN other words we get the exact line whenever b = - image41.pngimage41.png .

The y-intercept of the family of curves goes from - ∞           + while the x-intercept is always positive.

The solution of the 2x2 general system of equations is:

x + ay= a²        y= - image42.pngimage42.png + a

bx – y= image38.pngimage38.png         y= bx - image38.pngimage38.png

-image44.pngimage44.pngx + a = bx - image45.pngimage45.png          (b + image44.pngimage44.png)x = a + image45.pngimage45.png          x=  image47.pngimage47.pngx = image48.pngimage48.png

y= bx - image38.pngimage38.png        y = b (image49.pngimage49.png) - image50.pngimage50.png               y= a - image50.pngimage50.pngy= image52.pngimage52.png

Both equations are the same, when b = - image41.pngimage41.png . This implies that a = -  image38.pngimage38.png.

        Therefore, the first equation, y= - image42.pngimage42.png + a, becomes y = bx –  image38.pngimage38.png.

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