Simultaneous Linear Equations

This lesson explains how to use matrix methods to (1) represent a system of linear equations compactly and (2) solve simulataneous linear equations efficiently.

How to Represent a System of Linear Equations In Matrix Form

Suppose you have n linear equations with n unknowns. Using ordinary algebra, those equations might be expressed as:

A₁₁x₁ + A₁₂x₂ + A₁₃x₃ + . . . + A_1nx_n = y₁
A₂₁x₁ + A₂₂x₂ + A₂₃x₃ + . . . + A_2nx_n = y₂
A₃₁x₁ + A₃₂x₂ + A₃₃x₃ + . . . + A_3nx_n = y₃
. . .
A_n1x₁ + A_n2x₂ + A_n3x₃ + . . . + A_nnx_n = y_n

where

x_j is an unknown value
A_ij is the known coefficient of x_j in equation i
y_j is a known quantity in equation j

This set of equations can be expressed compactly in matrix form as follows:

Ax = y

where

x is an n x 1 column vector of unknown values x₁, x₂, . . . , x_n
A is an n x n matrix of the known coefficients A_ij
y is an n x 1 column vector of known values y₁, y₂, . . . , y_n

How to Solve Simultaneous Linear Equations Using Matrix Methods

Here is how to solve a system of n linear equations in n unknowns, using matrix methods.

• Express the set of n linear equations compactly in matrix form.

  Ax = y

• Premultiply both sides of the equation by A^-1, the inverse of A.

  A^-1Ax = A^-1y

• Since A^-1Ax = Ix = x, we know the following.

  x = A^-1y

Thus, as long as the inverse A^-1 exists, we can solve for x, the vector of unknown values. If the inverse does not exist, the set of equations does not have a unique solution.

Solving Simultaneous Linear Equations: An Example

To illustrate how to solve simultaneous linear equations using matrix methods, consider the following system of equations.

  x₁ + 2x₂ + 2x₃ = 1
2x₁ + 2x₂ + 2x₃ = 2
  2x₁ + 2x₂ + x₃ = 3

We want to solve for the unknown quantities: x₁, x₂, and x₃.

• Our first step is to express these equations in matrix form as Ax = y.

1 2 2 2 2 2 2 2 1		x1 x2 x3	=	1 2 3
A		x		y

• Next, we premultiply both sides of the equation by A^-1, the inverse of matrix A. This results in the following relationship.

  A^-1Ax = A^-1y

  Recall that we showed how to find the inverse of matrix A in a previous lesson.

• And finally, since A^-1Ax = Ix = x, we know that x = A^-1y. Thus,

x1 x2 x3	=	-1 1 0 1 -1.5 1 0 1 -1		1 2 3	=	1 1 -1
x		A-1		y

Thus, we have solved for the unknown quantities: x₁ = 1, x₂ = 1, and x₃ = -1.

Test Your Understanding

Problem 1

Consider the following system of linear equations.

  3x₁ + x₂ = 3
9x₁ + 4x₂ = 6

Using matrix methods, solve for the unknown quantities: x₁and x₂.

Solution

Our solution involves a three-step process.

• The first step is to express these equations in matrix form as Ax = y.

3 1 9 4		x1 x2	=	3 6
A		x		y

• Next, we premultiply both sides of the equation by A^-1, the inverse of matrix A. This results in the following relationship.

  A^-1Ax = A^-1y

  Recall that we showed how to find the inverse of matrix A in a previous lesson.

• And finally, since A^-1Ax = Ix = x, we know the following.

x1 x2	=	4/3 -1/3 -3 1		3 6	=	2 -3
x		A-1		y

Thus, we have solved for the unknown quantities: x₁ = 2 and x₂ = -3.