Augmented Matrices - 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:

First step is to multiply Row 1 by , and substract Row 2, and put the resut in Row 2.

Now we have two equations.

, and

Solve for y first, and then plug the result into the other equation.

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: