Definition: the augmented matrix for a system of linear equations is
the matrix of coefficients with all coefficients of x ₁ in the first column,
all coefficients of x ₂ in the second column, etc. and all constants b
in the last column, separated from the body of the matrix by a bar.

ex: the augmented matrix of the system

The three elementary row operations applied to matrices are:

1. Multiply a row through by a nonzero constant k.

2. Interchange 2 rows

3. Add a multiple of one row to another.

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Gaussian Elimination

To solve systems of linear equations, we do row-reduction or elimination on the augmented matrix to put it into reduced row-echelon form or row-echelon form.

To be in reduced row-echelon form a matrix must have these properties:

1) If a row does not consist entirely of zeros, the first nonzero entry
in the row is a 1. (called a leading 1)

2) If there are any rows consisting entirely of 0's,
they are grouped together at the bottom of the matrix.

3) In any two consecutive nonzero rows, the leading 1 in the lower row
is farther to the right than the leading 1 in the higher row.

4) Each column with a leading 1 has 0's everywhere else.

A matrix having properties 1, 2, and 3 is in row-echelon form.

Example 1

These matrices are in reduced row-echelon form.

These matrices are in row-echelon form.

Every matrix has a unique reduced row-echelon form.
Row-echelon form of a matrix is not unique.

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Solving Systems by Elimination

When we solve a system by reducing the augmented matrix to row-echelon form
we use back-substitution to solve for the variables.

Example 2

Say we're solving a system of 3 equations in 3 variables and row-reduction
leads us to this:

Using back-substitution we get: z = 5 from row 3
From row 2, we get that y + 6z = 2 and since z = 5, y = 2 – 6 (5) = –28.
From row 1, we get that x + 4y + 3z = 7, but we know y and z, so x = 104

So back-substitution changes the matrix row back into an equation
into which we substitute the known values to find the rest.

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Now let's do one where there are more variables than equations.
Our solution will require a parameter.

Example 3:

Suppose the row-reduced augmented matrix of a system looks like this:

The 2nd row indicates that y + 3z = 2, so if we assign a value for z
we'll be able to solve for x and y. This is the role of the parameter.

The system's solution then is:

z = t

y = 2 – 3t

and from row 1

x = 19 – 26 t ; where t

These are the parametric equations of a line in 3-Space. Once we choose a
value for t, a point is fully defined. (see lesson on "Lines and Planes in 3-Space")

Since there are more unknowns than equations, the system has infinite solutions.

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Order of Row-Reduction Procedure

When performing row-reduction on a matrix, we proceed in this way.

1) put a leading 1 in the a₁₁ position.
2) make all a_i1 entries = 0 (1st column)
3) put a leading 1 in the a_{2 j} position; j > 1 (2nd row, further right)
4) clear the jth column above and below this leading 1. etc.

Note: if we keep track of the row operations we perform left of the matrix,
it is easier to find mistakes when they happen -- and they will happen!

Note: oprow indicates the row of the matrix we're using to make changes.

Example 3

Row reduce this augmented matrix to solve the system.

, so x = –1, and y = 3.

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Now a system with 3 variables.

Example 4

Solve the system with row reduction:

This last matrix says that x = –3, y = 9 and z = –7

Note: We didn't follow orders with this one. We didn't establish our leading 1
in the first row until we'd used the leading "2" to eliminate the column entries below it.
Sometimes, this tactic makes things more efficient.
As a general rule however, we should first try to get a leading 1 in the a₁₁ position.

Reminder: A matrix row is an equation.

A matrix column is a variable.

Note: When solving a system with M variables and n equations
where M > n, we will use a maximum of (M – n) parameters depending on
how many rows of zeros there are in the row-reduced matrix.

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Practice

1) Solve these systems with elimination

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Solutions

1) a) Reduction Steps

1. r₂ – r₁ and r₃ – 4r₁.
2. (1/4) r₂ and (1/5) r₃.
3. r₂ + r₁ and r₃ – r₂.
4. r₁ + r₃.
The solution is x = 1, y = 3, z = 0.

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1) b) Reduction Steps

1. Switch r₂ & r₃.
2. r₃ – 2r₁ and r₄ – 3r₁.
3. r₁ – 2r₂ and r₃ + r₂ and r₄ – r₂ .
4. r₄ – r₃ .

We get as the last row in the reduced matrix.
This of course means that z = 0. From rows 1 through 3, we get:

w = 5/6

x = –1/6

y = 1/6

z = 0

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1) c) Reduction Steps

1. r₂ + r₁ and r₃ – r₁ and r₄ – 2r₁ .
2. r₃ + r₂ and r₄ – r₂.

We get as the last 2 rows in the reduced matrix
which means the system is inconsitent. There are no solutions.

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