Systems of Linear Equations

Standard Form of 2 Equations with 2 Variables

𝑎1𝑥 + 𝑏1𝑦 = 𝑐1

𝑎2𝑥 + 𝑏2𝑦 = 𝑐2

where 𝑎1, 𝑎2, 𝑏1 , 𝑏2 are coefficients of variables 𝑥 and 𝑦, respectively, and 𝑐1, 𝑐2 are constants

Set of Solutions

set of all ordered n-tuples of numbers that make each equation a true statement

Equivalent

two linear systems that have the same solution set

Solving with 2 Variables

basic approach - simplify the problem by eliminating one of the variables, and finding the other

• substituting

• subtracting

• equating

Substituting

Let us go back to our set of equations:

𝑥 + 2𝑦 = 5

3𝑥 + 9𝑦 = 21

In the first equation we can isolate 𝑥 to obtain: 𝑥 = 5 − 2𝑦

We can now substitute the top equation for 𝑥 into the second equation:

3 (5 − 2𝑦) + 9𝑦 = 21

We have just eliminated one of the variables, i.e., 𝑥, by substitution

15 − 6𝑦 + 9𝑦 = 21

3𝑦 = 6

𝑦 = 2

𝑥 = 5 − 2 ∙ 2 = 1

Solution <x, y> = <1, 2>

Subtracting

𝑥 + 2𝑦 = 5

3𝑥 + 9𝑦 = 21

Observe that the first equation’s truth remains if we were to multiply it by some constant, i.e., 3

3𝑥 + 6𝑦 = 15

3𝑥 + 9𝑦 = 21

If we subtract two true equations from each other we obtain another true equation.

3𝑥 + 9𝑦 − 3𝑥 − 6𝑦 = 21 − 15

3𝑦 = 6

𝑦 = 2

𝑥 + 2 ∙ 2 = 5

𝑥 = 1

Solution <x, y> = <1, 2>

Equating

𝑥 + 2𝑦 = 5

3𝑥 + 9𝑦 = 21

We can isolate 𝑥 in both equations:

𝑥 = 5 − 2𝑦

𝑥 = (21 − 9𝑦)/3 = 7 − 3𝑦

The variable 𝑥 is still unknown, but we know two facts about it

𝑥 = 5 − 2𝑦 = 7 − 3𝑦; by adding 3𝑦 to both sides of the last equation:

5 + 𝑦 = 7; by subtracting 5

𝑦 = 7 − 5 = 2

𝑥 = 5 − 2 ∙ 2 = 1

Solution <x, y> = <1, 2>

Solving with 2 Variables Graphically

Representing the set of solutions of a system of two linear equations:

𝑎1𝑥 + 𝑏1𝑦 = 𝑐1

𝑎2𝑥 + 𝑏2𝑦 = 𝑐2

• set of points in ℝ² satisfying both equations simultaneously

• the intersection of the two lines in ℝ²

Types of Solutions with 2 Variables

A system of linear equations in which at least one of the coefficient is 𝑎 or 𝑏 ≠ 0 has

• exactly one solution, or

• infinitely many solutions, or

• no solution

A system of linear equation is consistent if it has either one solution or infinitely many solutions

A system of linear equation is inconsistent if it has no solution

Solving with 3 Variables

Representing the set of solutions of a system of three linear equations:

𝑎1𝑥 + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

𝑎2𝑥 + 𝑏2𝑦 + 𝑐2𝑧 = 𝑑2

𝑎3𝑥 + 𝑏3𝑦 + 𝑐3𝑧 = 𝑑3

• set of points in ℝ³ satisfying all three equations simultaneously

• the intersection of the three planes in ℝ³

Solutions with 3 Variables

Solving with n Variables

Basic algebraic approach: simplify the problem by eliminating one of the variables, and finding the other

• substituting

• subtracting

• equating

Advanced systematic approach: Gauss-Jordan elimination algorithm

Coefficient Matrix

Let’s us have:

𝑥 + 2𝑦 + 3𝑧 = 6	1 ∙ 𝑥 + 2 ∙ 𝑦 + 3 ∙ 𝑧 = 6
𝑥 − 𝑦 + 𝑧 = 1	1 ∙ 𝑥 − 1 ∙ 𝑦 + 1 ∙ 𝑧 = 1
2𝑥 + 3𝑦 + 4𝑧 = 9	2 ∙ 𝑥 + 3 ∙ 𝑦 + 4 ∙ 𝑧 = 9

The coefficient matrix:

• 𝐴 ∈ ℝ^{𝑚 × 𝑛}, a rectangular array of real numbers, with 𝑚 rows and 𝑛 columns

• one row for each equation

• one column for the coefficients of each variable, i.e., 𝑥, 𝑦 and 𝑧

Augmented Matrix

Let’s us have:

𝑥 + 2𝑦 + 3𝑧 = 6	1 ∙ 𝑥 + 2 ∙ 𝑦 + 3 ∙ 𝑧 = 6
𝑥 − 𝑦 + 𝑧 = 1	1 ∙ 𝑥 − 1 ∙ 𝑦 + 1 ∙ 𝑧 = 1
2𝑥 + 3𝑦 + 4𝑧 = 9	2 ∙ 𝑥 + 3 ∙ 𝑦 + 4 ∙ 𝑧 = 9

The augmented matrix:

•  ∈ ℝ^{𝑚 × (𝑛 + 1)}

• one row for each equation

• one column for the coefficients of each variable, i.e., 𝑥, 𝑦 and 𝑧

• one final column for constant terms, after the vertical line

Gauss-Jordan Elimination Algorithm

start with a system of linear equations expressed as augmented matrix

perform row operations to transform the matrix in equivalent forms

• swap the position of 2 rows

• multiply a row by a non 0 constant

• add a multiple of one row to another row

end result - reduced row echelon form of the matrix

Reduced Row Echelon Form

The leading entry in a row of a matrix is the first non-zero entry in that row, starting from the left:

1. all nonzero rows are above any rows of all zeros

2. each leading entry of a row is in a column to the right of the leading entry of the row above it

3. all leading entries in each non zero row are 1

4. all entries in a column below leading entry are zero

5. all entries in a column above leading entry are zero (1, 2, 3 in row echelon form)

Forward Phase (left to right)

1. Obtain a pivot (a leading one) in the leftmost column

2. Subtract/add the row with the pivot from all rows below it, to obtain zeros in the entire column

3. Look for a leading one in the next column to the right and Repeat

Outcome: row echelon form

Backward Phase (right to left)

1. Find the rightmost pivot and use it to eliminate all numbers above the pivot in its column

2. Move one column to the left and Repeat

Outcome: reduced row echelon form