# Patterns Within Systems Of Linear Equations

1820 words (7 pages) Essay

4th May 2017 Mathematics Reference this

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The purpose of this report is to investigate systems of linear equations where the systems’ constants have mathematical patterns.

The first system to be considered is a 2 x 2 system of linear of equations:

In the first equation, the constants are 1, 2, and 3 in that order. It is observed that each constant is increased by 1 from the previous constant. Thus, the constants make up an arithmetic sequence whereby the first term ( U1 ) = 1, and the difference between each term ( d ) = 1. Hence, the general formula is Un = U1 + (n-1)(1) where n represents the nth term.

U2 = U1 + (n-1)(d) U3 = U1 + (n-1)(d)

2 = 1 + (2-1)(1) 3 = 1 + (3-1)(1)

2 = 2 3 = 3

In the second equation, the constants are 2, -1, and -4 in that order. It is observed that each constant is increased by -3 from the previous constant. Thus, the constants from this equation also make up an arithmetic sequence whereby U1 = 2 and d = -3. Hence the general formula is Un = U1 + (n-1)(-3).

U2 = U1 + (n-1)(d) U3 = U1 + (n-1)(d)

-1 = 2 + (2-1)(-3) -4 = 2 + (3-1)(-3)

-1 = -1 -4 = -4

To further investigate the significance of these arithmetic sequences, the equations will be solved by substitution and displayed graphically.

x + 2y = 3 2x – y = -4

x = 3- 2y y = 2x + 4

x = 3 – 2(2) y = 2(3 – 2y) +4

x = -1 5y = 6 + 4

y = 2

On the graph, both lines meet at a common point (-1,2) where x = -1 and y = 2.

The two linear equations have a solution of x = -1 and y = 2, proven analytically and graphically.

However, this pattern may be only specific to this 2 x 2 system of linear equations. Therefore, other 2 x 2 system of linear equations following the same pattern of having constants forming arithmetic sequences will be examined as well.

Another 2 x 2 system of linear equations to be considered is:

The constants of these equations are 3, 6, and 9, and 4, 2, and 0 with a difference of 1 and -2 respectively.

The equations were then re-written as:

And plotted on a graph.

The common point of both equations is (-1,2), with x being -1 and y being 2. Therefore the common point has been proven both analytically and graphically to be (-1,2).

Another example is:

The constants of these equations are -3, 1, and 5, and -2, -6, and -10 with a difference of 4 and -4 respectively.

The equations were then re-written as:

And plotted on a graph.

The common point is (-1,2). Thus it is both proven analytically and graphically that the common point is (-1,2).

Another example is:

The constants of these equations are 3, 2, and 1, and 2, 7, and 12 with a difference of -1 and 5 respectively.

The equations were then re-written as:

And plotted on a graph.

The common point is (-1,2). Thus it is both proven analytically and graphically that the common point is (-1,2).

Another example is:

The constants of these equations are 5, 12, and 19, and 1, -5, and -11 with a difference of 7 and -6 respectively.

The equations were then re-written as:

And plotted on a graph.

The common point is (-1,2). Thus it is both proven analytically and graphically that the common point is (-1,2).

From the examples of 2×2 systems of linear equations, a conjecture that could be derived is:

“The solution for any 2×2 system of linear equations with constants that form an arithmetic sequence is always x=-1 and y=2.”

The general formula of such equations could be written as:

Whereby represents the first term for the first equation and represents the first term for the second equation with a common difference of and respectively.

The equations are then solved simultaneously:

Therefore, it is proved that the solution for a 2×2 system of linear equations with constants that form an arithmetic sequence is always x = -1 and y = 2.

However, the possibility of a 3×3 system exhibiting the same patterns as the previous 2×2 systems examined has not been discussed. Hence, this investigation will extend to 3×3 systems as well.

Here is an 3×3 system:

The for the first equation is 3 and the is (5-3)= 2.

The for the first equation is 1 and the is (-4-1)=-5.

The for the first equation is 4 and the is (7-4)=3.

Gaussian Elimination method will be used.

Change R3 into 4R2-R3

Change R2 into 3R2-R1

Change R3 into 23R2-17R3

The third row/R3 has all 0 which means that there is no one unique solution but infinite solutions. Therefore, in R2

We will let where k is a parameter

To find other solutions, will be substituted in the other equation

The solutions to this 3×3 system of linear equations with the pattern of constants making up an arithmetic sequence are , , and where is a parameter.

Here is another 3×3 system:

The for the first equation is 2 and the is (3-2)= 1.

The for the first equation is 5 and the is (5-3)=-2.

The for the first equation is -3 and the is (4-(-3))=7.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

The third row is all 0. This indicates that there is no unique solution, but infinite solutions instead.

Assuming that whereby is a parameter,

The solutions to this 3×3 system of linear equations with the pattern of constants making up an arithmetic sequence are , , and where is a parameter.

Another:

The for the first equation is 4 and the is (-2-4)= -6.

The for the first equation is 1 and the is (5-1)=-4.

The for the first equation is 2 and the is (7-2)=5.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

The third row is all 0. This indicates that there is no unique solution, but infinite solutions instead.

Assuming that whereby is a parameter,

The solutions to this 3×3 system of linear equations with the pattern of constants making up an arithmetic sequence are , , and where is a parameter.

Here is another 3×3 system:

The for the first equation is 4 and the is (-4-4)= -8.

The for the first equation is 2 and the is (-1-2)=-3.

The for the first equation is 6 and the is (14-6)=8.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

The third row is all 0. This indicates that there is no unique solution, but infinite solutions instead.

Assuming that whereby is a parameter,

The for the first equation is 7 and the is (20-7)= 13.

The for the first equation is 20 and the is (3-20)=-17.

The for the first equation is 6 and the is (-5-6)= -11.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

Assuming that whereby is a parameter,

From these examples, a conjecture can be made. A 3×3 system of equations that have constants that form an arithmetic sequence will have infinite solutions that will be in the form of , , and where is a parameter.

This is proven by the general formula:

Being solved by using Gaussian elimination rule:

Change R3 into R3-R2

Change R2 into R2-R1

Change R3 into

Change R2 into

Change R3 into R3-R2

R3 has only zeroes/0. This means that there is no unique solution but infinite solutions instead.

Assume whereby is a parameter,

Through substitution,

The solutions for this 3×3 system are , , and , proving the conjecture true.

Other than systems of linear equations that contain arithmetic sequences, other types will be investigated.

Let’s consider this 2 x 2 system:

In the first equation, the constants 1, 2, and 4 make up a geometric sequence whereby the first term (U1) is 1 and each consecutive term is multiplied by a common ratio (r) which in this case is 2.

â†’ U2 â†’ U3

In the second equation, the constants 5, -1, and make up a geometric sequence whereby U1 = 5 and r = – .

â†’ U2 = â†’ U3

The equations can be rewritten in the form of as:

For the first equation, and .

For the second equation, and..

The relationship between and appears to be that one is the negative reciprocal of the other. In any case, more examples of similar linear equations will be needed to thoroughly investigate the patterns.

The equations will be solved by substitution:

Another example:

In the first equation, the constants 3, 12, and 48 make up a geometric sequence whereby the first term (U1) is 3 and each consecutive term is multiplied by r which in this case is 4.

â†’ U2 â†’ U3

In the second equation, the constants 3, -1, and make up a geometric sequence whereby U1 = 3 and r = – .

â†’ U2 = â†’ U3

The equations can be rewritten in the form of as:

For the first equation, and .

For the second equation, and..

The equations will be solved using substitution:

Another example:

In the first equation, the constants 7, 42, and 252 make up a geometric sequence whereby the first term (U1) is 7 and each consecutive term is multiplied by r which in this case is 6.

â†’ U2 â†’ U3

In the second equation, the constants 2, -1, and make up a geometric sequence whereby U1 = 2 and r = – .

â†’ U2 = â†’ U3

The equations can be rewritten in the form of as:

For the first equation, and .

For the second equation, and..

The equations will be solved by using substitution:

From observing all three systems, it is found that the relationship between and appears to be that one is the negative reciprocal of the other. But it can also be said that .

The general formula of such equations could be written as:

Whereby represents the first term for the first equation and represents the first term for the second equation with a common ratio of and respectively.

The equations are then solved simultaneously:

So is the result of one ratio subtracted from the other.

is the product of the common ratios from both linear equations.

The purpose of this report is to investigate systems of linear equations where the systems’ constants have mathematical patterns.

The first system to be considered is a 2 x 2 system of linear of equations:

In the first equation, the constants are 1, 2, and 3 in that order. It is observed that each constant is increased by 1 from the previous constant. Thus, the constants make up an arithmetic sequence whereby the first term ( U1 ) = 1, and the difference between each term ( d ) = 1. Hence, the general formula is Un = U1 + (n-1)(1) where n represents the nth term.

U2 = U1 + (n-1)(d) U3 = U1 + (n-1)(d)

2 = 1 + (2-1)(1) 3 = 1 + (3-1)(1)

2 = 2 3 = 3

In the second equation, the constants are 2, -1, and -4 in that order. It is observed that each constant is increased by -3 from the previous constant. Thus, the constants from this equation also make up an arithmetic sequence whereby U1 = 2 and d = -3. Hence the general formula is Un = U1 + (n-1)(-3).

U2 = U1 + (n-1)(d) U3 = U1 + (n-1)(d)

-1 = 2 + (2-1)(-3) -4 = 2 + (3-1)(-3)

-1 = -1 -4 = -4

To further investigate the significance of these arithmetic sequences, the equations will be solved by substitution and displayed graphically.

x + 2y = 3 2x – y = -4

x = 3- 2y y = 2x + 4

x = 3 – 2(2) y = 2(3 – 2y) +4

x = -1 5y = 6 + 4

y = 2

On the graph, both lines meet at a common point (-1,2) where x = -1 and y = 2.

The two linear equations have a solution of x = -1 and y = 2, proven analytically and graphically.

However, this pattern may be only specific to this 2 x 2 system of linear equations. Therefore, other 2 x 2 system of linear equations following the same pattern of having constants forming arithmetic sequences will be examined as well.

Another 2 x 2 system of linear equations to be considered is:

The constants of these equations are 3, 6, and 9, and 4, 2, and 0 with a difference of 1 and -2 respectively.

The equations were then re-written as:

And plotted on a graph.

The common point of both equations is (-1,2), with x being -1 and y being 2. Therefore the common point has been proven both analytically and graphically to be (-1,2).

Another example is:

The constants of these equations are -3, 1, and 5, and -2, -6, and -10 with a difference of 4 and -4 respectively.

The equations were then re-written as:

And plotted on a graph.

Another example is:

The constants of these equations are 3, 2, and 1, and 2, 7, and 12 with a difference of -1 and 5 respectively.

The equations were then re-written as:

And plotted on a graph.

Another example is:

The constants of these equations are 5, 12, and 19, and 1, -5, and -11 with a difference of 7 and -6 respectively.

The equations were then re-written as:

And plotted on a graph.

From the examples of 2×2 systems of linear equations, a conjecture that could be derived is:

“The solution for any 2×2 system of linear equations with constants that form an arithmetic sequence is always x=-1 and y=2.”

The general formula of such equations could be written as:

Whereby represents the first term for the first equation and represents the first term for the second equation with a common difference of and respectively.

The equations are then solved simultaneously:

Therefore, it is proved that the solution for a 2×2 system of linear equations with constants that form an arithmetic sequence is always x = -1 and y = 2.

However, the possibility of a 3×3 system exhibiting the same patterns as the previous 2×2 systems examined has not been discussed. Hence, this investigation will extend to 3×3 systems as well.

Here is an 3×3 system:

The for the first equation is 3 and the is (5-3)= 2.

The for the first equation is 1 and the is (-4-1)=-5.

The for the first equation is 4 and the is (7-4)=3.

Gaussian Elimination method will be used.

Change R3 into 4R2-R3

Change R2 into 3R2-R1

Change R3 into 23R2-17R3

The third row/R3 has all 0 which means that there is no one unique solution but infinite solutions. Therefore, in R2

We will let where k is a parameter

To find other solutions, will be substituted in the other equation

Here is another 3×3 system:

The for the first equation is 2 and the is (3-2)= 1.

The for the first equation is 5 and the is (5-3)=-2.

The for the first equation is -3 and the is (4-(-3))=7.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

Assuming that whereby is a parameter,

Another:

The for the first equation is 4 and the is (-2-4)= -6.

The for the first equation is 1 and the is (5-1)=-4.

The for the first equation is 2 and the is (7-2)=5.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

Assuming that whereby is a parameter,

Here is another 3×3 system:

The for the first equation is 4 and the is (-4-4)= -8.

The for the first equation is 2 and the is (-1-2)=-3.

The for the first equation is 6 and the is (14-6)=8.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

Assuming that whereby is a parameter,

The for the first equation is 7 and the is (20-7)= 13.

The for the first equation is 20 and the is (3-20)=-17.

The for the first equation is 6 and the is (-5-6)= -11.

The equations were put into matrix form and row reduction was done on the Graphic Design Calculator.

Assuming that whereby is a parameter,

From these examples, a conjecture can be made. A 3×3 system of equations that have constants that form an arithmetic sequence will have infinite solutions that will be in the form of , , and where is a parameter.

This is proven by the general formula:

Being solved by using Gaussian elimination rule:

Change R3 into R3-R2

Change R2 into R2-R1

Change R3 into

Change R2 into

Change R3 into R3-R2

R3 has only zeroes/0. This means that there is no unique solution but infinite solutions instead.

Assume whereby is a parameter,

Through substitution,

The solutions for this 3×3 system are , , and , proving the conjecture true.

Other than systems of linear equations that contain arithmetic sequences, other types will be investigated.

Let’s consider this 2 x 2 system:

In the first equation, the constants 1, 2, and 4 make up a geometric sequence whereby the first term (U1) is 1 and each consecutive term is multiplied by a common ratio (r) which in this case is 2.

â†’ U2 â†’ U3

In the second equation, the constants 5, -1, and make up a geometric sequence whereby U1 = 5 and r = – .

â†’ U2 = â†’ U3

The equations can be rewritten in the form of as:

For the first equation, and .

For the second equation, and..

The relationship between and appears to be that one is the negative reciprocal of the other. In any case, more examples of similar linear equations will be needed to thoroughly investigate the patterns.

The equations will be solved by substitution:

Another example:

In the first equation, the constants 3, 12, and 48 make up a geometric sequence whereby the first term (U1) is 3 and each consecutive term is multiplied by r which in this case is 4.

â†’ U2 â†’ U3

In the second equation, the constants 3, -1, and make up a geometric sequence whereby U1 = 3 and r = – .

â†’ U2 = â†’ U3

The equations can be rewritten in the form of as:

For the first equation, and .

For the second equation, and..

The equations will be solved using substitution:

Another example:

In the first equation, the constants 7, 42, and 252 make up a geometric sequence whereby the first term (U1) is 7 and each consecutive term is multiplied by r which in this case is 6.

â†’ U2 â†’ U3

In the second equation, the constants 2, -1, and make up a geometric sequence whereby U1 = 2 and r = – .

â†’ U2 = â†’ U3

The equations can be rewritten in the form of as:

For the first equation, and .

For the second equation, and..

The equations will be solved by using substitution:

From observing all three systems, it is found that the relationship between and appears to be that one is the negative reciprocal of the other. But it can also be said that .

The general formula of such equations could be written as:

Whereby represents the first term for the first equation and represents the first term for the second equation with a common ratio of and respectively.

The equations are then solved simultaneously:

So is the result of one ratio subtracted from the other.

is the product of the common ratios from both linear equations.

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