1. Systems of Linear Equations
Gaussian Elimination
Types of Solutions
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

2. A linear equation is an equation that can be written in the form:
The coefficients ai and the constant b can be real or complex numbers.
A Linear System is a collection of one or more linear equations in the same variables. Here are a few examples of linear systems:
Any system of linear equations can be put into matrix form:
The matrix A contains the coefficients of the variables, and the vector x has the variables as its components.
For example, for the first system above the matrix version would be:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

3. Infinitely many solutions
Before diving into larger systems we will look at some familiar 2-variable cases. If the equation has two variables we think of one of them as being dependent on the other. Thus we have only one independent variable. A one-dimensional object is a line, so solutions to these two-equation systems can be thought of as the intersection points of two lines. We will generalize this concept when dealing with larger systems.
Consider the following sets of equations:
Unique Solution
These two lines intersect in a single point.
No Solution
Here the lines are parallel, so never intersect.
In this case we call the system inconsistent.
Infinitely many solutions
The two lines coincide in this case, so they have an infinite number of intersection points.
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4. Infinitely many solutions
Here are some 3-variable systems.
Each equation represents a plane (a 2-dimensional subset of ℝ3).
We are looking for intersections of these planes.
Unique Solution
If the coefficient matrix reduces to the identity matrix there will be a unique (constant) solution to the system.
No Solution
If the system is inconsistent there will be no solutions.
In this case there will be a contradiction that appears during the solution process.
Infinitely many solutions
If, after row reduction, there are more variables than nonzero rows, the system will have a family of solutions that can be written in parametric form.
We will use a procedure called Gaussian Elimination to solve systems such as these.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

5. Elementary Row Operations
Here is our first 3x3 system. To start the Gaussian Elimination procedure (also called Row Reduction), form the Augmented Matrix for the system. This is simply the coefficient matrix with the vector from the right-hand-side next to it. The procedure will use “Elementary Row Operations” to give us a matrix in “Reduced Echelon Form”.
Augmented Matrix
Elementary Row Operations
Add a multiple of one row to another row
Switch two rows
Multiply any row by a (nonzero) constant
Performing any of these operations will not change the solutions to the system. We say that systems are “row-equivalent” when they have the same solution set.
Echelon Form
All nonzero rows are above any rows of all zeros.
Each leading entry of a nonzero row is in a column to the right of the leading entry in the row above it.
All entries in a column below a leading entry are zero.
To be in Reduced Echelon Form, there are two more requirements:
The leading entry in each nonzero row is 1.
Each leading 1 is the only nonzero entry in its column.
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6. Here is our first 3x3 system.
The first row-reduction step is to get zeroes below the 1 in the upper left.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

7. Here is our first 3x3 system.
The first row-reduction step is to get zeroes below the 1 in the upper left.
For the next step we want to get a zero in the bottom-middle spot. For hand-calculations I like to keep entries integers as much as possible, so I would do 2 steps here-multiply row 3 by 2, then add (-3) times row 2. This can be done together:
Next step is to multiply (-1) through row 3, then use row 3 to get zeroes in column 3.
After dividing row two by 2, the last step is to add row 2 to row 1.
This matrix is in “Reduced Echelon Form” (REF)
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8. Here is the reduced matrix
Here is the reduced matrix. Notice that the left part has ones along the diagonal and zeroes elsewhere. This is the 3x3 Identity Matrix. In this case the solution is staring us in the face.
The solution represents the common intersection point of the three planes represented by the equations in the system.
Unique Solution
If the coefficient matrix reduces to the identity matrix there will be a unique (numerical) solution to the system.
The solution to this system is a single point: (1,4,5)
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9. Here is the next system. The basic pattern is to start at the upper left corner, then use row operations to get zeroes below, then work counterclockwise until the matrix is in REF.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

10. Here is the next system. The basic pattern is to start at the upper left corner, then use row operations to get zeroes below, then work counterclockwise until the matrix is in REF.
At this point you might notice a problem. That last row doesn’t make sense. It might help to write out the equation that the last row represents.
It says 0x+0y+0z=3.
Are there any values of x, y and z that make this equation work? (the answer is NO!)
This system is called INCONSISTENT because we arrive at a contradiction during the solution procedure. This means that the system has no solution.
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For Campus Learning Assistance Services at UCSB

11. This is the reduced matrix (actually we could go one step further and get a zero up in row 1). Notice that we got a row of zeroes in the left part of the augmented matrix. Since the augmented side is nonzero we got a contradiction.
In cases where the augmented side is also zero (a row with all zeroes) we can ignore that row of the reduced matrix, as it yields no information.
No Solution
If the system is inconsistent there will be no solutions.
In this case there will be a contradiction that appears during the solution process.
This line is the intersection of a pair of the planes
This line is the intersection of a different pair of the planes, and it is parallel to the other line, so there are no intersection points.
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12. This is two equations in 3 unknowns. The free variable is z.
Here is another system. Work through the row-reduction procedure to obtain the REF matrix.
This time there is a row of zeroes at the bottom. However there is no contradiction here.
The bottom row represents the equation 0x+0y+0z=0. This is always true!
Continue with row reduction by dividing row 2 by -3, then get a zero in row 1.
Reduced Echelon Form
Now that we have the reduced matrix, how do we write the solution? A good place to start is to write the system of equations represented by the reduced matrix:
This is two equations in 3 unknowns. The free variable is z.
This means that z can be any value, and the system will have a solution.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

13. Infinitely many solutions
To write down the solution, we can rearrange the equations so that they are solved in terms of the free variable (z).
Here I have written the solution in vector form (re-naming the parameter ‘t’ instead of ‘z’). For each value of t there is a corresponding point on the line.
Notice that is similar to the typical equation of a straight line : y=mx+b.
Infinitely many solutions
If, after row reduction, there are more variables than nonzero rows, the system will have a family of solutions that can be written in parametric form.
This system has a 1-parameter solution: it is a line in ℝ3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB