PRACTICAL MATHEMATICS

Simultaneous Linear Equations

Contents

1. Basic Examples with Two Unknowns
2. Gaussian Elimination
3. General Theory
4. Example Using the General Theory

Basic Examples with Two Unknowns

In the simultaneous equations

a₁₁x₁ + a₁₂x₂ = b₁
a₂₁x₁ + a₂₂x₂ = b₂,

x₁ and x₂ are the unknowns, a₁₁, a₁₂, a₂₁ and a₂₂ are given numbers called the coefficients, and b₁ and b₂ are given numbers on the right sides of the equations. A pair of values for x₁ and x₂ which make both equations true at the same time is called a solution of the equations.

The examples below show that the pair of equations may have

• one unique solution,
• infinitely many solutions,
• or no solution.

The number of solutions depends on the values of a₁₁, a₁₂, a₂₁, a₂₂, b₁ and b₂. If the equations have one unique solution, a quick method of calculating it is by means of the formulas:

x₁ = (a₂₂b₁ - a₁₂b₂)/(a₁₁a₂₂ - a₁₂a₂₁)
x₂ = (-a₂₁b₁ + a₁₁b₂)/(a₁₁a₂₂ - a₁₂a₂₁)

The expression (a₁₁a₂₂ - a₁₂a₂₁) is called the determinant of the system of equations. If the determinant is zero, then the above formulas cannot be used and the equations have either infinite many solutions or no solution.

If b₁ and b₂ are both zero, the pair of equations is called homogeneous. If b₁ and b₂ are not both zero, the pair of equations is called non-homogeneous.

Example 1: Non-Homogeneous Equations with One Solution

x1 + x2 = 1 x1 - x2 = 0

There is one unique solution x₁ = 1/2, x₂ = 1/2, at the point where the two graphs intersect.

Example 2: Homogeneous Equations with One Solution

x1 + x2 = 0 x1 - x2 = 0

There is one unique solution, the trivial solution x₁ = 0, x₂ = 0, at the origin where the two graphs intersect.

Example 3: Non-Homogeneous Equations with Infinitely Many Solutions

x1 + x2 = 1 2x1 + 2x2 = 2

There are infinitely many solutions because the two equations are represented by the same graph. Every point on the graph is a solution.

Example 4: Homogeneous Equations with Infinitely Many Solutions

x1 + x2 = 0 2x1 + 2x2 = 0

There are infinitely many non-trivial solutions because the two equations are represented by the same graph through the origin. Every point on the graph is a solution.

Example 5: Inconsistent Equations.

x1 + x2 = 1 x1 + x2 = 0

There is no solution because the two graphs are parallel; they have no point of intersection.

Gaussian Elimination

Gaussian elimination is a practical method of solving simultaneous linear equations. It is easy to see from the results of this method whether the equations are consistent or inconsistent, and whether consistent equations have one unique solution or infinitely many solutions.

The following examples show the method.

Example: Three Equations and Four Unknowns

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0.6x₁ + 1.5x₂ + 1.5x₃ - 5.4x₄ = 2.7
1.2x₁ - 0.3x₂ - 0.3x₃ + 2.4x₄ = 2.1

Use the first equation to eliminate x₁ from the second and third equations as follows:

• Subtract (0.6)/3 times the first equation from the second equation.
• Subtract (1.2)/3 times the first equation from the third equation.

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0x₁ + 1.1x₂ + 1.1x₃ - 4.4x₄ = 1.1
0x₁ - 1.1x₂ - 1.1x₃ + 4.4x₄ = -1.1

Use the second equation to eliminate x₂ from the third equation.

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0x₁ + 1.1x₂ + 1.1x₃ - 4.4x₄ = 1.1
0x₁ + 0x₂ + 0x₃ - 0x₄ = 0

No more eliminations are possible. The last equation is reduced to the true numerical statement

0 = 0.

This shows that the equations are consistent. Since the number of equations is less than the number of unknowns, there are infinitely many solutions. For example, we can give any numbers we choose to x₃ and x₄. Then the second equation gives x₂, and (using this value of x₂) the first equation gives x₁.

Example: Inconsistent Equations

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0.6x₁ + 1.5x₂ + 1.5x₃ - 5.4x₄ = 2.7
1.2x₁ - 0.3x₂ - 0.3x₃ + 2.4x₄ = 2.0

These equations are the same as the equations in the above example except that the right side of the third equation is 2.0 instead of 2.1. After the Gaussian elimination method has been done we get

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0x₁ + 1.1x₂ + 1.1x₃ - 4.4x₄ = 1.1
0x₁ + 0x₂ + 0x₃ - 0x₄ = -0.1

The last equation is reduced to the false numerical statement

0 = -0.1

This shows that the equations are inconsistent, and that the equations have no solution.

EXERCISES

Solve the following sets of simultaneous linear equations.

General Theory

Suppose we have m simultaneous equations with n unknowns:

a11x1	+	...	+	a1nxn	=	b1
...		...		...		...
am1x1	+	...	+	amnxn	=	bm

Inconsistent Equations

If the Gaussian elimination method produces a false numerical statement 0 = c, where c is not zero, then the equations are inconsistent and there is no solution.

One Unique Solution

If the equations are consistent and the Gaussian elimination method produces exactly n non-zero equations, then the simultaneous equations have one unique solution.

Infinitely Many Solutions

If the equations are consistent and the Gaussian elimination method produces r non-zero equations, where r < n, then r unknowns are determined in terms of the remaining n - r unknowns. These remaining n - r unknowns can be given any chosen values.

The number r of non-zero equations produced by the Gaussian elimination method is called the rank of the system of equations.

Homogeneous Equations (b₁ = 0, ..., b_m = 0)

If the equations are homogeneous, then there is a trivial solution x₁ = 0, ..., x_n = 0. There are non-trivial solutions if and only if r < n; then there are exactly n - r linearly independent non-trivial solutions

(x₁⁽¹⁾, ..., x_n⁽¹⁾), ..., (x₁^(n-r), ..., x_n^(n-r)).

The general solution of a homogeneous system with r < n is a linear combination of these solutions:

x1	=	c1x1(1)	+	...	+	cn-rx1(n-r)
...		...		...		...
xn	=	c1xn(1)	+	...	+	cn-rxn(n-r)

where the coefficients c₁, ..., c_n-r can be given any values we choose.

Example Using the General Theory

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0.6x₁ + 1.5x₂ + 1.5x₃ - 5.4x₄ = 2.7
1.2x₁ - 0.3x₂ - 0.3x₃ + 2.4x₄ = 2.1

As shown above, Gaussian elimination reduces these equations to:

3x₁ + 2x₂ + 2x₃ - 5x₄ = 8
0x₁ + 1.1x₂ + 1.1x₃ - 4.4x₄ = 1.1

A particular solution, obtained by putting x₃ = 0 and x₄ = 0, is

x₁ = 2, x₂ = 1, x₃ = 0, x₄ = 0.

The complementary homogeneous equations with the same coefficients as the original equations are:

3x₁ + 2x₂ + 2x₃ - 5x₄ = 0
0.6x₁ + 1.5x₂ + 1.5x₃ - 5.4x₄ = 0
1.2x₁ - 0.3x₂ - 0.3x₃ + 2.4x₄ = 0

Gaussian elimination reduces these equations to:

3x₁ + 2x₂ + 2x₃ - 5x₄ = 0
0x₁ + 1.1x₂ + 1.1x₃ - 4.4x₄ = 0

Since r = 2 and n = 4, we have r < n. Since n - r = 2, it follows that there exist exactly 2 linearly independent solutions. For example, by putting x₃⁽¹⁾ = 1, x₄⁽¹⁾ = 0, and x₃⁽²⁾ = 0, x₄⁽²⁾ = 1, we get:

x₁⁽¹⁾ = 0, x₂⁽¹⁾ = -1, x₃⁽¹⁾ = 1, x₄⁽¹⁾ = 0,
x₁⁽²⁾ = -1, x₂⁽²⁾ = 4, x₃⁽²⁾ = 0, x₄⁽²⁾ = 1.

This gives the general solution of the homogeneous equations as follows:

x₁ = c₁.0 + c₂.(-1) = -c₂
x₂ = c₁.(-1) + c₂.4 = -c₁ + 4c₂
x₃ = c₁.1 + c₂.0 = c₁
x₄ = c₁.0 + c₂.1 = c₂

The general solution of the complementary homogeneous equations is also called the complementary solution of the original non-homogeneous equations. The general solution of the original non-homogeneous equations is:

General Solution = Particular Solution + Complementary Solution

In this example the general solution of the original non-homogeneous equations is therefore:

x₁ = 2 - c₂
x₂ = 1 - c₁ + 4c₂
x₃ = 0 + c₁
x₄ = 0 + c₂

By R. H. B. Exell, 2001. King Mongkut's University of Technology Thonburi.
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