**PATTERNS WITHIN SYSTEM OF LINEAR EQUATION**

A system of linear equation is basically dealt with in the algebra unit. It is a collection of the linear equations involving variables of the same set in the in the equations that are involved. For example a 2×2 system of linear equations includes:

x + 2y=10

3x + 4y=15

Here in both the cases the equations only involve two variables that is x and y and no other variable is included.

In the example of a 33 system of linear equations it mostly includes the variables x, y and z for example;

2x + y-z =11

x- 2y + 2z =-2

3x-y+2z =5

Where only the three variables are involved

There are also various properties of the patterns of the linear systems. We will start with the consistency property. If the systems of the equations have common solutions, then they are said to be consistent. This therefore means that graphically the lines should be straight lines. The independence property is also termed as the linear independence. The systems of equations are usually independent since to start with, they are derived algebraically from others. For example the system 3x+4y =9 and 6x +8y =18.

There are different ways of solving the systems of linear equations that includes;

- The elimination of variables
- The substitution of variables technique
- The row reduction method
- The crammers’ rule
- The matrix method

In the mathematical field, the general linear equation in the x and y is Ax+By=C where both the A and B in the equation are not zeros. The y-intercept in the line is the y-coordinate of that point where graphically, the non-vertical line that is drawn either manually or graphically intersects the y-axis. Also, the x-intercept is the point where the non horizontal line crosses the x-axis. Therefore the most general equation for a line with slope m and the y-intercept passing through b as the y intercept is written as y= mx + b.

Therefore, one can easily find the slope and at the same time the y-intercept of any line. For example finding the slope and the y-intercept for 4x+5y=40

Solution: first and foremost, solve the equation for y to put it in the slope intercept format

4x+5y=20 5y=20-4x

y=4-4/5x

y=-4/5x+4 therefore the slope m=-4/5 and the y intercept is b=4

Consider this 2×2 system of linear equations

4x+3y=7

3x-2y=9

When we examine our first equation 4x+3y=7, there is a pattern in the constants of the equations used. Here 4 is the constant associated with the variable x and it therefore precedes the variable x. Also 3 is a constant that is preceding the variable y and the equation results to 7. In the second equation, 3x-2y=9, the constant 3 precedes our variable x and the constant -2 precedes the variable y making the equation to result to 9. It is also clear that in the two equations, the constants both have a difference of one.

Solving the equations simultaneously, we first multiply the first equation by 3 and then multiply the second equation by 4 in order to eliminate the variable x and solve for the variable y. The equation then becomes;

12x+9y=21

12x-8y=36

17y=-15

Therefore solving the equation yields y=-15/17. Putting the value of x in any of the solution to obtain the value of x;

4(x) +3(-15/17) = 21

X=41/17

Graphically the system of equation is solved as

This is first done by putting the two equations in the form of y=mx+b. The solution of the equations is by observing the point of intersection of the two lines that are plotted graphically.

In this system of equation the solution therefore is (41/17, -15/17)

Consider this 2×2 system of linear equations

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

The two equations are linear because the unknowns only appear to the first power, no unknown in the denominator of a fraction is in the equations and there are no products of unknowns.

Therefore, the most general linear equation is

a_{11}x_{1 }+ a_{12}x_{2 }+………+ a_{1n}x_{n}=b_{1}

a_{21}x_{1} +a_{22}x_{2}+_{………………}+a_{2n}xn=b_{2}

a_{m1}x_{1}+ a_{m2}x_{1} +………..+a_{mn}x_{n}=bm_{ m }

With unknowns x_{1}, x_{2}…..x_{n} and coefficients a_{1, }a_{2}……a_{n }.

In x+2y=3, the constant is 3 and the unknowns are x and y whereby x= 3-2y and y= (3-x) ÷2.

The gradient of the linear equation is -1/2 and the y- intercept is 3/2. The gradient is negative therefore it is negatively sloped.

In 2x-y =-4, the constant is -4 and the unknowns are x and y where x= (y-4) ÷2 and y=2x+4.

The gradient of the line is 2 and the y- intercept is 4. The gradient is positive and therefore positively sloped.

Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area

[1 2: 3] and [2 -1: -4]

The inverse is

The solution therefore is;

*1 2: x =3*

*2 -1: y =-4*

*1/5 2/5 3 =x*

*2/5 -1/5 -4 =y*

The graphic calculator here was used during this step to solve the matrix equation, normally if the equation is Ax= B then the solution is x= A^{-1}B

X=-1

y=2

The first function from the graph is sloped from left to right that is it is negatively sloped since the gradient is negative and the second equation is sloped from right to left since the gradient is positive.

From the graph the solution of the equations is x and y2. This is read directly from the graph where the two lines intersect.

In short, the solution to the system of equation is unique in that there is only one solution set to the system of equations and the solution satisfies the individual equations in the system of equations.

Therefore, when x=-1 and y=2. Then

-1+ (2×2) =3 and (2×-1) – 2 = -4 which satisfy the equations that are given.

- Another example of linear equations is

x+2y =3

3x-5y=9

This is a system of equations since it contains more than one equation. The solution set to the system of linear equation is the set of numbers n and m such that if we let x=n and y=m then we will obtain the result of the right hand side of the equation.

For instance ax+by =c, if x=n and y=m then we obtain the result c given a and b are known constants.

Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area

1 2: x =3

3 -5: y =9

The inverse of the equation is

-5/11 -2/11

-3/11 1/11

The solution therefore is,

-5/11 -2/11 3 =x

-3/11 1/11 9 =y

Therefore xand y

The graph of the two equations is as shown below

This is so since (3 1) + (2 0) =3 and (33)-(50) = 9 as proven from the equation.

Now consider the 2 2 system below

x2y=4

5x-y=1/5

In the first equation, x2y=4 f(x) = x/8. Therefore (0, 0)

In the second equation, 5x-y=1/5 then f(x) = 5x-1/5. The gradient is 5 and the y intercept is -1/5

The matrix of the equation therefore is

[1/8 -1: 0]

[5 -1: 1/5]

Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area

-39/8 39/8: 0 = x

-195/8 39/64: 1/5 = y

- Consider the following two by two system of equation

y + 2x=7

y + x/2=3

In the first equation, the line is negatively sloped since y= f(x) = -2x+7. The gradient is -2 which means that the change in x compared to the change in y is -2. The y intercept is 7 and therefore when x=0, y=7

In the second equation, the equation line is also negatively sloped. The equation y= f(x) =-x/2 +3. The gradient is -1/2 and the y- intercept is 3 meaning when x=0, y=3

Therefore the solution to the equation is x=8/3 and y=5/3

The graph of f(x) =-2x+7 and f(x) =-x/2+3 are as follows:

From the graph both the linear equations are negatively sloping but that of f(x)= -2x+1 is steeper than that of –x/2 + 3.

The equations above are linear equations that results to linear curves and therefore two equations are enough to solve the equations. This two equations result into a square matrix.

In the multiplication of matrices for instances, if A is an np matrix and B is a pm, then AB is the product of A and B denoted by AB and AB will be an nm matrix. That is AB exists if and only if number of columns of A is equal to the number of rows of B and that it should be noted that ABBA therefore the matrices do not commute. Therefore in solving the equation for example Ax=B, to find the values of x, the inverse of A is first found then multiplied by B. That is x= A^{-1}B

If A is a square matrix, we can find another matrix B known as the inverse of the matrix such that AB=BA=I. The inverse however can be a right inverse or a left inverse most commonly denoted as A^{-1.}

If AB=BA=I then A^{-1}= B. Therefore if B exists then A is said to be invertible and non singular matrix. If B does not exist then A is said to be singular matrix. For the solution to be found in the equations, the matrix involved should be non singular.

Theorem

If A is a non singular matrix then A^{-1} is unique

Proof

Let A^{-1 }=B then AB=BA=I

Suppose B is not unique, then there exist C such that CB

Then CA=AC=I

But B=IB therefore (AC) B= (CA) B and thus C (AB) =CI =C

Therefore B=C, a contradiction that B is unique

An example of invertible matrix can be solved by looking at the following question.

Solve the equations -4x-2y=8 and 6x+3y=12

The matrix of the equation is a 22 matrix and therefore the equation can written in the augmented form as shown

Any matrix is said to be in reduced row echelon form if it satisfies the following conditions

- Any row of all zeros appear at the bottom
- If a row does not consist of all zeros then its first non zero entity is called a leading 1 and it is one
- If any two successive rows the leading 1 of the lower is further to the right of the leading 1 of the highest row
- If a column contains a leading 1 then all the other entries are zero

However, each of the unknown variables can be solved using the graphical calculator in the matrix calculation area

But the inverse of the matrix does not exist since in fact the determinant is zero (-43) – (6-2) =0. Therefore there is no solution to the equation above.

Graphically, the lines to the equations are parallel and never intersect therefore there is no solution to the problems

From the graph, it is clear that the two lines are parallel and are never to intersect and therefore this means that the equations do not have a solution.

In some cases, the system will have many solutions in the algebraic sense, however geometrically, the lines will collide and look like there is only one line and therefore all the points along that line are indeed solutions to the equation. For example the equation -2x+y=8 and -4x+2y= 16 matrix to the equation is

One of the solutions is x=0 and y=-4 and many other solutions.

The matrix does not have an inverse as shown by the graphical calculator in the matrix calculation area since the determinant of the matrix is zero.

This means that the solution of the equation is not one hence the equation has many solutions. In general, given any systems of linear equation with two unknown solution, the two lines will graphically intersect at one point. The point of intersection is the solution to the systems of linear equations. Also, the lines can be parallel to each other meaning that the system does not have any solution and finally the lines can collide and the solution to the system is not unique that is there are many solutions to the system of linear equations.

The system with many solutions can be presented in the graph below using the equation given above. Note that:

-4x+2y=16 2 {(-2x+y) =8} and therefore one equation is a multiple of the other which basically means that the equation is more or less the same.

Remember that a system such as 2x-3y=7 and x+7y=11 can be written in the form

=

It is normally represented as Am=b where A= , b= and m=

If b=0, then the system Am= b has m=0 as a trivial solution but if A^{-1} exists then m=0 is the only solution to the system.

There is also the possibility of graphing those equations with piecewise defined functions. For instance there are functions such as

|x| =

In our example, we will graph one of the most commonly used piecewise defined functions.

f (x) =

In this case, the entire function is considered as one function in who’s the domain is the real numbers.

APPLICATION OF LINEAR FUNCTIONS

There are times when the solutions for the complicated functions cannot easily be obtained. This lead to the use of the linear equations that is the equations that are to only one degree to be used in the approximation of the complicated functions since they gives some little bit of accuracy and the linear functions are easy to work with. This is basically known as linearization. This is mostly used in conjunction with the differential functions. Here if the function, normally denoted as f is differentiable at x=a, then in this case the approximation function denoted as L(x) = f (a) +f’ (a) (x-a) is now what is known as the linearization of the function f at a.

For example we will try to find the linearization of the function f(x) = at x=0

The above graph is now the linearization of the function at x=o and x=3.

We now know that f’(x) =1/2(1+x)^{-1/2}. We will therefore see that f (0) =1 and also that f’ (0) =1/2.

Therefore this concludes that the linearization will therefore be:

L(x) = 1+1/2(x-0) = 1+x/2

These are some of the general applications of the linear equations/functions and many other that are dealt with at the higher level of the course work

Theorem in the solution of system of equation

If A is invertible, then there is only one solution to Am=b which is the unique solution

Proof

Let w be any solution such that wA^{-1}b

That is Aw=b but since A is invertible A^{-1} exists that is Aw=b

Therefore multiplying both sides to the left with A^{-1} we have A^{-1}Aw = A^{-1}b

I w = A^{-1}b

w = A^{-1}b which is a contradiction and therefore A^{-1}b is the only solution to the system

Next there are equations that are to the second degree and the linear equations are used to find the gradients at particular points through the use of the tangent line and the normal lines to the equations that are being considered in this case. These equations mostly include the parabolas and other quadratic equations among others. Though our main interest is not the parabolas and such equations, the linear equations are particularly used here to serve various mathematical purposes. The parabola for instance is a set that usually consist of all points in a plane that is equal in distance sense from a point that is given and also a given line. Mostly the parabolas will have a graph of equations of the form y= ax^{2}+bx+c.

We will for instance plot a graph of y=x^{2}. In this case the graph is a simple graph that is curved in u shape. But mathematically, we may want to find the gradient of the graph at particular points. We will therefore use the current technology for graph plotting to plot both the graphs as shown;

The tangent line is used to find the gradient of the curve at that particular point.

The graph shown is a curve with the equation y=x^{2}. The axis of the parabola is the y-axis that is it is the axis of symmetry.lso, the vertex of the parabola as seen from the graph is at the origin. The parabola is seen to open upwards when the values of the constant are positive and increasing and open up downwards if the values of the constant are negatively increasing.

Now if we consider the 33 matrix system, there are 3 variables that are involved, we will concentrate on the variables x, y and z. For instance, let as consider the matrix below

2x+y-z=11: Here the constants are 2 that precede the variable x, 1 that precedes the variable y and -1 that precedes the variable z.

The system can be solved using the usual matrix method, the elimination method or the use of a three dimension matrices. When we deal with the matrix method the graphing calculator here is used to find the inverse of the matrix. The system of equations can be basically being written as;

M X=A

Using the graphing calculator to find the inverse of the matrix will yield

X= M^{-1}A

X=

And therefore the solution to the equation becomes

X =

In this type of system, there are also the possibility of obtaining a unique solution, the; possibility of many solution and the option of no solution. The possibility of many solutions or no solution is as a result of having a singular matrix that is a matrix with a zero determinant. For example looking at the following system of solution

x+2y+3z=4

4x+6y+8z=10

2x+y=-1

The determinant of the matrix is zero and therefore there can be the case where there are many solutions and graphically in a three dimension graph, the lines are common or the case where there are is no solution and the lines are parallel to each other.

We can also use our technology to create a family of linear equations that are usually similar in characteristics. On the same set of the axis, we usually display the equations and evaluate them mathematically. The family of curves will include several lines which usually have a wide range of equations. This can be represented as;

The family of linear equations above all have different gradients fro negative to zero to positive.

In a 33 matrix, the solution can also be obtained geometrically and algebraically. This is so because the graph of the equations can be plotted in the graph especially with the current technologies and calculators and it can be done algebraically through various methods which include the elimination methods and the current modern methods. Therefore the 33 matrix can be dealt with in the same manner as the 22 matrix.

There are many ways of proving mathematical theorems and terms such as the contradiction method, proving by induction and many others in the above matrix we have used the contradiction method. In the 33 matrix, we are going to basically see how to prove by induction the conjectures that are involved. Conjectures in mathematics are some of the propositions and they are easily not disapproved since they are believed to be true

For instance the sequence a_{n}= n (n-1) is the sequence such that

a_{1}= 1*0

a_{2}= 2*1

:

:

a_{n}= n*(n-1)

When we sum up the sequence of the first n numbers we obtain a series and therefore s_{n}= a_{1}+a_{2}+……+a_{n. }Therefore;

S_{1}=0

S_{2}=2

S_{n}= S_{n-1}+a_{n}

Next, the difference between successive sums is made until the constant term in the series is obtained so long as the nth term n>0,

This will result to a polynomial of the third degree in order for the constant terms to be obtained in that the equation for the series will therefore be S_{n }= Ax^{3}+Bx^{2}+Cx+D where A,B,C and D are constants that are and xⱤ

Now replacing x in the equation with the natural numbers 1,2,3,4,5….. we get the

A+B+C+D=0

8A+4B+2C+D=2

27A+9B+3C+D=8

64A+16B+4C+D=20

This is a four equation system since there are four different unknown variables. Therefore we will use the graphing calculator to find the solutions to the unknown variables

1 1 1 1: 0

8 4 2 1: 2

27 9 3 1: 8

64 16 4 1: 20

We will find that A=1/3, B=0, C=-1/3 and D=0

S_{n}=1/3 x^{3}-1/3 x

The graph for the equation is therefore as follows

Proving the equation by the induction method therefore will be (for n>0);

For n=1: =1/3*1^{3 –} 1/3* 1 =0

For n=3: =1/3*3^{3} -1/3*3 =8

For n=5: =1/3*5^{3} – 1/3*5 =40

Therefore we can assume that the equation is true for all values of natural number that is n>0,

We therefore assume that the equation is true for n=k

Therefore for n=n+1,

S_{k+1}= S_{k}+ (k+1)*k

=1/3k^{3 }-1/3k +k^{2 }+k

=1/3(k+1)^{3} -1/3(k+1)

Since the expression is true for n=k+1 is true, the equation is true by induction.

In the mathematical sense, a function of a polynomial p is normally written as p(x) =a_{n}x^{n}+a_{n-1}+….+a_{1 }x+ a_{0}. In this case the n are non negative integers and the a’s are the coefficients of the polynomial itself. Usually all the polynomials have the domain of (-, ). In this case we can say that the linear functions themselves are polynomials of degree one while the quadratic functions are polynomials of the second degree and so on. As with our 33 matrix, the polynomial involved was a cubic function of the third degree. For instance the polynomial y=8x^{4}-14x^{3-9x}2+11x-1

Linear Algebraic Equations

A teacher is looking for the best option in purchasing school supplies for a classroom. Company A is offering a discount for every dollar amount spent; Company B is offering a higher discount for every dollar spent above $20. Determine which company will offer a better price based upon the dollar amount the teacher spends on the school.

In this scenario, it mostly involves the computation of the purchase of the school inventories at a cheaper price. Inventory generally is the stock of raw materials, work in progress units, finished goods, consumables and spare parts being held in store at a given time period. There are different kinds and groups of inventories that includes; movement inventories which are inventories on transit from one point to another, safety stock or the buffer stock which are the inventories that must always be maintained in the store so as to meet the unexpected demand, cyclical inventory, anticipatory inventory and the decoupling inventory. However, in this scenario we will focus on how to purchase inventory while at the same time using the mathematical knowledge to reduce inventory related costs. Here, the teacher is looking for the best option in purchasing school supplies for a classroom and therefore the best option is the option with reduced costs. The customer also has to ensure that though the goods are purchased at a cheaper price, they are of the best and desirable quality. The supplier of the goods should also be in a position to supply goods to the customer when they are needed both in the short term period and in the long term period and in time as to the date of the specifications. Therefore the customer has to look deep into these needs before making the decision on where to make their orders.

The hypothetical customer, the teacher in this case has to make an informed decision based on questions such as how many units to order at that time, how often should the school supplies be made, how many orders are to be placed in that particular year and this is mainly done to reduce cost.

In this scenario therefore we will focus mainly on two cost options that are for Company A which is offering a discount for every dollar spent. This is where a constant rate of discount for every dollar spent. This is where a constant rate of discount is offered irrespective of the number of units purchased and it is commonly known as a single discount. We will assume that the unit price of each product that is to be purchased is $5 and that the discount for every dollar spent is 5 percent (5%). In this scenario, the discount offered is for every dollar that will be spent and no conditions as to the amount and the limit of expenditure.

The second option is for Company B which is offering a higher discount for every dollar spent above $20. This therefore guarantees the teacher discount after spending $20 in the purchase of school supplies which will be a much higher discount than that of the purchase of the goods worth $20. In this case the teacher will get a discount similar to that of the single discount up to the expenditure of $20 and later the discount is increased accordingly. We will therefore assume that the unit purchase price is $5 and that every dollar spent to the expenditure of $20 is 5%, with more dollars spent, the discount increases to 7.5%. In our scenario, the demand should be known in advance with certainty and will remain constant within the relevant range.

The algebraic equations to represent the cost of each option are:

Company A: Offers a discount for every dollar spent

Here there are many cost related to the purchase of the school supplies which includes the purchase cost, the ordering cost, the holding cost and in some cases the shortage cost. However we will only focus on the purchase cost and ignore all the other related cost inorder to come up with the required linear equations.

Let’s assume further that the teacher purchases x units of the school supplies

Unit price = $5

Discount =5%

Let the total cost=y

Total purchase cost =$5 * (100% -5%) *x

Total purchase cost = $5 *0.95 *x

= 4.75x

Company B: Offers a higher discount of 7.55 for every dollar spent above $20.

The teacher here should know that for the first $20 spent, the discount is 5% and above the expenditure of $20, the discount increases to 7.5%. This will probably lure customers desire to purchase more but we will try evaluating the two equations. The equation for company B is therefore as follows

Let’s assume that the teacher purchases x units of the school supplies

The discount for the first 4 purchases of the school supplies =5% i.e ($20/5)

Unit price =$5

Let the total cost=y

The discount for the purchase of more than 4 =7.5%

Total purchase cost = ($5*4*0.95) + ($5*0.925)(x-4)

Total purchase cost = $19+4.625x –$18.5

=$0.5+4.625x

The equation of company A is used since the company only offers a single discount for all the purchases that are made by the customer. Therefore the discount will be distributed equally. For Company B, there is a constant in the algebraic equation since in the purchase of the first $20 items, the discount is 5% that is it is constant and since the customer has to purchase more than this to gain the discount of 7.5% then that part of the equation will vary with the extra units purchased.

The solution to the equation can be done through several ways such as elimination method, substitution method or the graphical method; the equations are normally written as

y= 4.75x

y= 0.5+4.625x

We are going to solve the equation using the substitution method. Since in the first equation y=4.75x, we will substitute this to the second equation.4.75x= 0.5+4.625x and we therefore solve the equation mathematically. In this equation, the solution is x=4. Where the total costs will be the same. But with the increase in the purchase of the school supplies, the total cost will be higher for the purchase related to Company A than that of the purchase from Company B. Also, with the decrease in the purchase of the school supplies, the cost purchases from Company A are less than that of Company B.

The graph of the two scenarios can be represented as follows. Though the cost associated in the two scenarios are close, there is a negligible difference as a result of the discounts.

It is therefore correct to conclude that if the teacher is in need of less than four units of purchase, it is advisable to purchase the school supplies from Company A, if the teacher wants to purchase 4 units of item, this can be done from any company and if it is more than 4 company, it is cost effective to purchase from Company B.

At the lower levels of purchase, presence of discount appears attractive for the company with a single discount for any unit of purchase made. That is, there is no constant related to the purchase of goods in the algebraic equation.. However beyond a certain level of purchase, taking up a discount results into a net increase in total cost in the company using a single discount method. In the other case for company with an increased discount after purchase of some discount, the purchase of many items become cost effective in this company. Therefore the teacher should ensure that he or she takes up the least quantity required to qualify for the highest discount in order for the total cost to be less than that of company A. Since the principle of discount states that only the least quantity required to qualify for the discount should be purchased.