Gamma Function

Further Algebra

determinants : simultaneous equations : partial fractions: limits


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Determinants

Introduction

A determinant is a collection of numbers set in rows and columns and enclosed by a pair of square brackets []. Each distinct value is called an element. Determinants have the same number of rows as columns. A determinant with n rows and columns is called an nth order determinant.

For example the determinant below is a second order determinant:

Second Order Determinant

This is an example of a third order determinant:

Third Order Determinant

Evaluating Determinants

A determinant has a value. To evaluate a second order determinant, diagonal elements are multiplied together and subtracted as shown below.

Evaluating Second Order Determinants

Example 1: Evaluate the determinant below.

Example
Using the formula gives a value of 11:

Evaluation

Example 2: Evaluate the determinant below.

Example

Using the formula gives a value of -7:

Evaluation

A third order determinant can be broken down into three second order determinants.

Evaluating Third Order Determinants

Each second order determinant can then be evaluated individually and multiplied by the term outside because:

Factors

Example 3: Evaluate the determinant below.

Example

The evaluation is in two stages via three smaller determinants:

Evaluation

A determinant with its rows or columns all zero evaluates to zero.

Zero Determinant

The evaluation of a determinant can be simplified. If a row or column has a common factor, that factor can be removed outside the determinant if all the rows / columns are then divided by that factor. This can lead to great simplifications of the arithmatic. In the determinant below, 6 is removed from column 1, then 2 is removed from row 1.

Simplifying

Solving Simultaneous Equations

Determinants can be used to solve simultaneous equations. Consider the following pair of simultaneous equations in two unknowns (x and y):

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

In the Introduction to Algebra, simultaneous equations were solved by manipulating them until one variable was left.

This is the method for solving simultaneous equations by using determinants.

Example 4: Solve the two simultaneous equations below:

x + y - 3 = 0 and x - y - 7 = 0

Create and evaluate the three determinants.

Evaluate D(n)

Evaluate D(x)

Evaluate D(y)

Apply the formula and replace with the values of the determinants.

Solution

Equating the fraction containing x with the fraction containing only numbers gives:

Solution

Equating the fraction containing y with the fraction containing only numbers gives:

Solution

The solutions are therefore x = 5 and y = -2.

Three simultaneous equations in three unknowns (x, y, z) can be solved in simliar way. Four third order determinants are produced by using the coeficients and numbers as above to give D(n), D(x), D(y) and D(z). The formula for solving the simultaneous equations is:

Solving

Example 5: Solve the three simultaneous equations below:

4x - 2y + 3z + 6 = 0
x + 2y - z - 7 = 0
2x + 3y + 2z - 4 = 0

First create the four determinants. D(n) contains the coeficients of x, y, and z. This has a value of 33.

Solving

D(x) contains all the coeficients and numbers except those for x. Its value is -33.

Solving

D(y) contains all the coeficients and numbers except those for y. Its value is 66.

Solving

D(z) contains all the coeficients and numbers except those for z. Its value is also 66.

Solving

Finally, apply the formula and replace with the values of the determinants.

Solving

By equating each fraction that contains an unknown with the fraction containing the numbers we can show that

x = 1, y = 2, z = -2.

Simultaneous equations of higher order with more unknowns can be solved just as easily.


Partial Fractions

Consider a fractional function of the form shown below where a, b, c, d and e are numbers:

Fractional Function

In many problems it is useful to be able to express this in the form of partial fractions like so

Partial Fraction

where A, B and f, g are numbers. The expressions x + f and x + g are the two factors of the original function's denominator (the lower part).

The method for converting fractional functions to partial fractions involves factorising the denominator of the original fraction and then evaluating the numbers A and B. An example is the easiest way of showing the method.

Example 6: Convert the fractional function below into partial fractions.

Partial Fractions Example

Firstly we factorise the denominator as shown below. We are using the identity sign (three lines) rather than the equals sign (=) because we are changing the format without affecting any values.

Partial Fractions Example

We then separate the function into two new fractions each with one of the factors as a denominator. The numbers in the numerators (the top parts), A and B, are not known at this stage and will need to be calculated.

Partial Fractions Example

In the next stage we add the fractions using the normal rules of fractional algebra. The numerator of the first fraction (A) is multiplied by the denominator of the second fraction (x + 2). The numerator of the second fraction (B) is multiplied by the denominator of the first fraction (x - 1). The combined numerator is then expanded and the various terms grouped together in terms of x and constants.

We are attempting to recreate the original fraction but with the A's and B's in the numerator:

Partial Fractions Example

We can now compare the original fractional function with the new version that we have created.

Partial Fractions Example

The left hand side is identical to the right hand side. This means that the coeficients of x can be equated on both sides and the numbers can also be equated. This gives a pair of simultaneous equations with two unknowns (A and B).

Partial Fractions Example

These equations can be easily solved to give A = 1 and B = 3, so

Partial Fractions Example


Limits

What is a Limit?

The idea of a limit is a strange one. It has been used in mathematics for thousands of years to give values to things that cannot be calculated exactly.

For example, consider the infinite series below:

Series

The series has an infinite number of terms. The sum of the series (how much the total is) can be rewritten using the summation symbol (the Greek capital letter, sigma Σ) as follows

Sum of Series

The symbol Σ means add up all terms like the following. In this case the terms are one over 2 raised to the power of n (1/2n). The notation n= 0 below the Σ means begin when n is 0. The infinity symbol above the Σ means carry on until infinity. So the meaning of the notation above is:

Add up all terms of the form 1/2n starting with n at 0 up to infinity.

Of course, we cannot do this sum because there are an infinite number of things to add. However, it can be shown that the more terms we take the closer the sum approaches to the value 2. We can use the idea of a limit to say that:

In the Limit, as n approaches infinity, the sum of the series approaches 2.

There is a concise way of writing this statement:

Limit Notation

The following limit

Limit

means

In the Limit, as n approaches infinity, 1 / xn approaches 0.

A final example

Limit

This one means

In the Limit, as x approaches 0, 1 / x approaches infinity.

The Derivative of x2

Limits are very useful mathematically. One example is in the differential calculus. We have already introduced the idea that the derivative of x2 is 2x. This will now be proved.

Consider the function

y = x2

This function has a value (y) for each value of x. To find a value for y, just square the value of x. When we find a derivative we are looking to see how this function varies at a given value of x. We can approximate this by adding a small amount to x and seeing how it affects y.

A small addition to x is given the symbol Δx. Similarly, a small addition to y is given the symbol Δy. The amounts have to be small otherwise we are not finding a variation at a particular point. When we do the addition we obtain:

y + Δy = (x + Δx)2

We can expand the right hand side

y + Δy = x2 + 2xΔDx + (Δx)2

Now subtract y from both sides

Δy = x2 + 2xΔx + (Δx)2 - y

Remember that y = x2

Δy = x2 + 2xΔx + (Δx)2 - x2 = 2xΔx + (Δx)2

Divide both sides by Δx gives

Δy / Δx = 2x + Δx

The expression Δy / Δx is pronounced delta y by delta x. It is a measure of how y varies when x is changed by a small amount. The smaller we make Δx, the closer the expression approaches the rate of change at a given point, the instantaneous rate of change. We cannot just set Δx to 0 because we cannot divide by zero.

But we can use limits. We can allow Δx to approach 0 and see what the limit is. This can be expressed:

Limit of Delta y by Delta x

This limit is the derivative. It has a value 2x. The derivative of x2 is 2x. This can be expressed, using the dy/dx notation as

dy/dx as a Limit

L'Hôpital's Rule

L'Hôpital's Rule allows limits of fractional functions to be found if they can be differentiated. If the two functions are f(x) and g(x) and if their separate derivatives are F(x) and G(x) respectively, then L'Hôpital's Rule states:

L'Hopital

If the derivatives do not give a limit then, the process can be repeated. The use of this rule is best shown in an example.

Example 7: Find the limit of (1 - cos x) / x2 as x approaches 0.

We are looking for the following limit.

L'Hopital

It is not easy to see what the limit is because as x approaches zero, both top and bottom approach zero. The limit cannot be determined from this expression. However, according to L'Hôpital's Rule, we can find the derivatives of top and bottom to give a new expression; if that limit exists (for x approaching 0) it will equal the limit for the above expression.

The derivative of 1 - cos(x) is sin(x). the derivative of x2 is 2x. Therefore we are looking for the following limit:

L'Hopital

This is no better because the limit is 0 / 0 as x approaches zero.

We can continue and differentiate top and bottom again. The derivative of sin(x) is cos(x). The derivative of 2x is 2. Our third limit is therefore:

L'Hopital

This does have a limit that is easy to see. As x approaches zero, cos(x) approaches 1. As x approaches zero, the denominator remains constant at 2.

Therefore the limit of this expression is 1/2. By L'Hôpital's Rule, 1/2 is also the limit of the original function when x approaches zero.

© 2004, 2009 KryssTal


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