Systems of Equations - College Algebra

1) Rewrite both the equations to tidy up and make sure all constant terms are isolated on one side of the equation. This will make it easier to write the augmented matrix.

2) Write the augmented matrix corresponding to the system.

Quick Refresher:

The identity matrix has the form,

We want to perform row operations until we have reduced the augmented matrix down to the form,

where the column vector represents what is left over from all the row operations performed to reduce the coefficient matrix down to the identity matrix. These will be the solution for and .

3) Now we can perform a succession of row operations to reduce to the desired form,

We can start by cancelling out the in the second row by subtracting the first row from the second row to obtain:

Now we need a to take the place of the in the top row, so divide row 1 by :

Now we need a to take the place of the in the second row, so divide the second row by ,

Now we need a to replace the remaining in the first row. Subtract 2-times the second row from the first row,

4) Write the Solution Set:

From the augmented matrix, we see that

Now perform row operations to find x and y.

1. Switch Row 1 with Row 2:

2. Multiply Row 1 by-2 and add it to Row 2:

3. Multiply Row 3 by 13 and multiply Row 2 by -1. Add the two rows for a new Row 2:

4. Multiply Row 3 by -5 and add it to Row 1. Multiply Row 2 by -2 and add it to Row 1:

If, during the process of Gauss-Jordan elimination, any row becomes an all zero-row, it ceases to be a factor in determining the solution set. We must examine the other rows of the matrix.

Remember that each entry to the left of the divider represents a coefficient and each entry to the right represents a constant. Therefore, the two nonzero rows in the matrix

represent the equations

and

,

respectively.

The equations can be solved for and in terms of :

and

Therefore, can assume any real value, with and dependent on ; this is a dependent system, with infinitely many solutions.

The red flag in this augmented matrix is the bottom row, which has 0's as all of its entries to the left of the divider and a nonzero entry to the right. Any time this happens during the process of Gauss-Jordan elimination, this signals that the system of equations has no solution.

If, during the process of Gauss-Jordan elimination, any row becomes an all zero-row, it ceases to be a factor in determining the solution set. We must examine the top row of the matrix alone.

Remember that each entry to the left of the divider represents a coefficient and each entry to the right represents a constant. Therefore, the top row in the matrix

represents the equation

,

or, equivalently,

Therefore, can assume any real value, and . There are therefore infinitely many solutions to this equation.

When applying the Gauss-Jordan method to solve a two-by-two linear system, the objective is to use row operations to form an augmented matrix of the form

.

If this happens, then there is one and only one solution to the system, represented by the ordered pair . The augmented matrix given,

is in this form, and there is one solution, .

When applying the Gauss-Jordan method to solve a three-by-three linear system, the objective is to use row operations to form an augmented matrix of the form

,

with the solution of the system.

Once the initial augmented matrix

is set up, the first step should always be to get a 1 in the upper left position; this is usually done by multiplying every element in Row 1 by the reciprocal of the first element in that row. Since the element in that position is , every element in Row 1 is multiplied by reciprocal . This step can be written using the notation .

Begin by constructing an augmented matrix for our system of equations:

Switch rows 1 and 2:

Add row 2 and twice row 1, to row 1:

Divide row 1 by 5:

Subtract 3 times row 1, from row 2:

Divide row 2 by 2:

Solution:

Plug the respective values of x and y into both equations to verify the solution:

Construct an augmented matrix for our system of equations:

Swap rows 1 and 3:

Subtract 2 times row 1 from row 2:

Subtract row 1 from row 3:

Add row 2 and three times row 3, to row 3:

Divide row 3 by 10:

Add row 3 to row 1:

Subtract row 3 from row 2:

Divide row 2 by 3:

Solution:

Plug the respective values of x, y, and z into all equations to verify the solution:

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

Then, add that to the third equation so that the y and z terms are eliminated. You will get .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

We can use this z value to find y

So the solution set is x = 1, y = 2, and z = –5/3.

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

Step 4: solve for z.

Step 5: solve for y.

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

To solve the system of linear equations, first isolate the equations with and to help solve for . This will include substitution.

Now that is known it can be substituted back into the third equation to solve for .

Lastly, substitute into the first equation and solve for .

Therefore, the solution to this system is

To solve the equations we want to try to cancel out some of the letters to reduce the equations. We can do this by adding different equations to each other. If we add:

we get a new equation,

.

We can also add

to get .

So we can rearrange the new equations to get

and substitute these into the original second line equation:

Then we can solve for b and c:

Add the first and third equations together, to eliminate z:

+

This yields the following:

Divide out a 2, to simplify the equation:

Plug this value back into the first equation:

Solve for z:

Add the first and second equations together to eliminate y:

+

This yields the following:

Divide out a 3 to simplify the equation:

Plug in the value for z:

Solve for x:

Plug the values for x and z into the first equation:

Simplify:

Solve for y:

Make sure the values for x,y, and z satisfy all of the equations:

Solution:

Add the first two equations to eliminate the y and z-variable.

Using the value of x, with the first and third equations, we will need to eliminate the z-variable to solve for y.

Multiply the top equation by three.

Subtract both equations.

Divide both sides by two.

The answer is:

The given system has as many variables as equations, which makes it possible to have exactly one solution.

One way to identify the solution set is to use Gauss-Jordan elimination on the augmented coefficient matrix

Perform operations on the rows, with the object of rendering this matrix in reduced row-echelon form.

First, a 1 is wanted in Row 1, Column 1. This is already the case, so 0's are wanted elsewhere in Column 1. Do this using the row operation:

Next, a 1 is wanted in Row 2, Column 2. This is already the case, so 0's are wanted elsewhere in Column 2. Do this using the row operations:

Next, a 1 is wanted in Row 3, Column 3.Perform the operation

Now, 0's are wanted in the rest of Column 3:

The matrix is in the desired form and can be interpreted to mean that the system has one and only one solution - . This makes the system consistent and independent.

rearranges to

and

, so

In the second equation, you can substitute for from the first.

Now, substitute 2 for in the first equation:

The solution is

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .