- Level: International Baccalaureate
- Subject: Maths
- Word count: 1473
Systems of Linear Equations. Investigate Systems of linear equations where the system constants have well known mathematical patterns.
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Introduction
Mathematical Investigation
Investigate Systems of linear equations where the system constants have well known mathematical patterns.
Part A:
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= -1
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).
Graph
Here are some examples using the same pattern:
Example 1:
Middle
c(a+ b) – cay – cby + cay + day= a(c + 2d) ca + 2cb – cby + day= ca + 2ad 2cb – cby + day= 2ad
ax= a + 2b – (a + b)(2) ax= a + 2b – 2a – 2b
day – cby= 2ad – 2cb y(ad – bc)= 2(ad – bc) y= 2 y=2
a(1 – 2) x= -1
a
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.
Conclusion
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= - as for b = 2; for a = - as for b = 5; etc. IN other words we get the exact line whenever b = - .
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= - + a
bx – y= y= bx -
-x + a = bx - (b + )x = a + x= x =
y= bx - y = b () - y= a - y=
Both equations are the same, when b = - . This implies that a = - .
Therefore, the first equation, y= - + a, becomes y = bx – .
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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