Using Elimination to Solve Linear Equations

There are different types of equations in mathematics. There can be different methods to solve these equations. One can use solve by elimination method to solve the equations. The equations can be linear and also quadratic in nature. Sometimes they are cubic in nature. This method is used to solve a set of linear equations. If the numbers of variables to be found out are ‘2’, then a set of equations must be given to find them.

A set of equations means two equations must be given. The number of equations can be deciding factor in finding the solution. In this method an unknown variable is eliminated first. Then the value of the second variable can be easily found out. Once the value of the first variable is found out then this value can be substituted in any one of the given equations and the second variable can be easily found out. This is one of the easiest techniques that can be used to find the unknowns in linear equations. For solving equations having three unknowns three equations must be given, otherwise it is not possible to find the unknowns. This is a prerequisite to find the unknowns.

The ‘solve by elimination technique is used when a set of linear equations are to be solved. This technique can be very useful and simple to use. There are various techniques available to solve linear equations. This is one of the simplest methods and can easily understood and used.

This is why it is most commonly used by the students. It is also easy to cross-check the answers. The solution obtained must satisfy the given equations. The process of solving by elimination can make the task of solving of linear equations very easy.

Once solve by the elimination method is learnt, the other methods to solve the linear equations can also be learnt. This is the basis for the other methods. So, one must thoroughly understand this method. Here elimination refers to the fact that a variable is first eliminated in this method. Once a variable is eliminated the second variable can be easily found out. Then this value helps in finding the other variable.

This can be done by using the principle of substitution. The value found out is substituted back into the equation and the value of the other variable is easily found out.

Diagonal Matrix Properties

A (mxm) matrix is a square array of numbers which has equal number of rows and columns here denoted by ‘m’. The elements of a square matrix that start from the left corner towards the right corner like a diagonal is called the Main Diagonal of a Matrix. For example, a 3x3 matrix given by
A=[-1  3  4]
 [2   1  0]
 [7  0   5]
The elements [ -1,1,5] form the elements of the main diagonal. Now consider A=(aij) to be a square matrix then the elements in aii is the diagonal of a matrix.  A square matrix in which all elements except the elements of the leading diagonal are zero is called a Matrix Diagonal. In this all the non diagonal elements are zero. For instance, a square matrix given by A=[aij]nxn to be a diagonal-matrix , aij is equal to zero wherever ‘i’ is not equal to ‘j’. Example of a diagonal-matrix can be given as follows, a 3x3 matrix given by
A=[2    0    0]
     [0   -1    0]
     [0    0    5]

Let us now take a look at the Diagonal Matrix Properties:
• If A and B are two diagonal matrices of the same order then, A+B and AB are also diagonal matrices.  Also AB=BA, that is they commute
• A square matrix is diagonal only if it is triangular and normal
• If A is a diagonal-matrix then the adjugate of A is also a diagonal matrix
• Any diagonal matrix is symmetric
• It is a scalar matrix when all the entries of the diagonal are equal
Determinant Diagonal Matrix
The determinant of a 2x2 square matrix A=[a    b] can be defined as |A|=ad-bc. It is denoted by det(A)
    [c    d]
Let A=[aij] be a nxn matrix where n is greater than or equal to two. We can define determinant,
|A|= a11C11 + a12C12+……..+a1nC1n which is equal to summation(i= 1 to n)[a1iC1i]. If A is a nxn diagonal matrix then |A| is the product of the elements in its main diagonal.
Let A = [a11       0       0……     0]
             [0         a22     0 ……   0]
     [0          0      a33……    0]
     [..         ..      ..     ..]
     [..         ..      ..     ..]
     [0          0        0…. .ann]

be a nxn diagonal matrix, then the determinant of this matrix can be given by a simple formula, det (A)=|A|= pi(i= 1 to n)[aii]
Consider the following example,
A=[2   0    0]
      [0   5    0]
      [0   0    1]

then det(A)=|A|= (2)(5)(1)=10
Note that if any of the elements of the main diagonal is zero then determinant would also be zero.

Process of Solving Inequalities

In mathematics both equalities and inequalities are present. The equalities give rise to equations. In equations the ‘equal to’ sign is used. There are various methods to solve the equations. The best and the easiest method must be used in order to arrive at the answer quickly and efficiently.

The concept of solving inequalities by addition and subtraction is very easy to understand and use. The concept can be very easy to understand if one is thorough with the concept of basic arithmetic operations like addition and subtraction.

The concept of inequality can be easily understood with the help of examples. The examples can help in explaining the concept very clearly.

In solving inequalities using addition or subtraction, the basic arithmetic operations of addition and subtraction are used. The inequality is called an inequality because the terms on the left side will not be equal to the terms on the right hand side. This is the basic reason for the term being called an inequality. This is the basic difference between the equality and inequality. In equality the terms on the left side will be equal to the terms on the right hand side.

The process of transposing will be used to solve the equations. Even in the case of inequalities the same process or principle is used. This is very simple principle. The terms on one side of the equation are moved to the others side; while doing so the sign of the term changes.

The positive term becomes the negative term and the negative term becomes the positive term. The multiplication becomes division and division becomes multiplication. So, one must be very careful while transposing the terms.

An inequality is formed when a term may be less on side or the other side. This is nothing but a side is surplus and the other side contains slack variable. So, the surplus has to be reduced to convert the inequality into equality.

But the same method can be used to solve both types of problems. Once the solutions are obtained for these types of problems it is necessary to check the feasibility of the solutions. All the solutions obtained might not satisfy the given equations. So, one has to be careful in selecting the final solutions. These solutions which are selected must satisfy the given equations. In case of inequality the solution must satisfy the inequality.

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Linear Programming Simplex Method

In Linear Programming Simplex Method is a method used for problem solving. This method is used for a problem of the form:
Minimize or maximize : c.x
Subject to
Ax=b, xi ≥0  
X is the variables of the problem (x1,x2…) while c is coefficients of the objective function (c1,c2,…). A is a matrix and b are constants≥0.
To solve the linear programming problems we must follow standard form which is achieved as:
A new variable should be added for all those variables whose upper bound is other than zero such that the new variable is the difference between original variable and upper bound.
For example: if x≥3 then introduce a new variable y such that
Y=x-3 and x=y+3.
For rest of the inequality constraints, new variable known as slack variable is introduced to remove inequalities. Like:
X+2y≤3 and –x+3y≥2 is replaced as:
X+2y+z=3 and –x+3y-z=2 such that z≥0.
Let us solve some problems using The Simplex Method

Question) Use Simplex Method to solve the following:
P=3x+4y subject to:
x+y≤4
2x+y≤5
x≥0,y≥0

Solution)  Since we have two constraints, we will introduce 2 slack variables p and q:
x+y+p=4
2x+y+q=5
We rewrite our objective function as −3x−4y+P=0 and thus system of equations become:
x+y+p=4
2x+y+q=5
−3x−4y+P=0
This gives initial simplex table:
X y p q P
1 1 1 0 0 4
2 1 0 1 0 5
-3 -4 0 0 1 0

Find the column with the most negative entry among x,y,p,q and P (here this is −4). Find pivot row by dividing each entry in the constant column by the entry in the corresponding in the pivot column. In this case, we get 4/1 as the ratio for the 1st row and 5/1 for the ratio in the 2nd row. The pivot row is the row corresponding to the smallest ratio which is 4 in this case. So our pivot element is in the 2nd column, 1st row =1.Now, perform the following row operations to convert the pivot column to a unit column:
R2→R2−R1
R3→R3+4R1
So, simplex table is changed to:
X Y p q P
1 1 1 0 0 4
1 0 -1 1 0 1
1 0 4 0 1 16
The variables are given the value in the constant column in the row where a value 1 is in the unit column. All variables above a non-unit column is given 0 value. So y=4, p=1, P=16, x=0, and q=0.
Thus, maximum occurs when x=0, y=4 and the maximum value is 16.
This is how Simplex Method Solver works.

Change of base formula

In logarithms, many times we encounter with the problem when there is a logarithm whose value is not known to us for that particular base, but we know its value for some other base.
For such cases we work by changing the base of logarithms by using Logarithmic Change of Base Formula which is easy to implement.

The Change of Base Formula for logarithm:
log a (x) = log b (x) / log b (a)
Here ‘a’ is base of logarithm of x before change and, b is base of logarithm of x after changing the base.
So above Log Change Base Formula can be used when you need to change base of logarithm from ‘a’ to b provided that the new base is useful to you, otherwise changing a base will not be useful.
Let us see proof of Logarithms Change of Base Formula and then we will go through some examples:

Proof:
We know that Raising ‘a’ (base of log a (x)) with the power of log a (x) gives x:
 x = a^(〖log〗_a  (x))………………………………………….(1)
Similarly, raising b with the power of log b (a) gives a:
a = b^(〖log〗_b  (a))        …………………………………………(2)
When we replace a in (1) with b^(〖log〗_b  (a)) given in equation (2), we get:
x = a^(〖log〗_(a ) (x)) =〖〖(b〗^(〖log〗_b  (a)))〗^(〖log〗_a  (x)) =  b^(〖log〗_b  (a)  〖* log〗_(a ) (x))     ………(3)
Applying log b() on both sides of equation (3) we get:
log b (x) = log b 〖(b〗^(〖log〗_b  (a)  〖* log〗_(a ) (x)))    
Using the log power rule we get:
log b (x) = log b (a)*log a(x)
Log a (x) = log b (x) / log b (a)
Now let us see some examples using the above formula:

Example 1) Change the base of following logarithm to 2: log12 64.
Solution) Using change of base formula this can be evaluated as:
log12 (64) = log2 (64)/log2 (12)

Example 2) Solve the following by changing base to e: log10ee
Solution) log10 ee = loge ee / loge 10
= e/ln 10 (loge 10 is written as ln 10)
= e/2.302   (as ln 10 =2.302; use calculator for this)

Example 3) Solve log9 27.
Solution) here it is difficult to find the value directly so we will change the base to 3 as 9 and 27 is square and cube numbers of 3 respectively.
Log 9 27 = log 3 27 / log3 9
= log3 33 / log3 32
= 3/2   (As loga bc = c(loga b) and also loga a = 1; these are other properties of logarithms)