## Practice problems for back substitution.

### First practice problem for back substitution.

We will introduce practice problems for back substitution, from Stewart’s book College Algebra. Two practice problems number 25&27.

1-For the first practice problem #25 for back substitution. A matrix is given in the row-echelon form. a) write the system of equations for which the given matrix is the augmented matrix. b) use back substitution to solve the system. The detail of the given matrix for practice problem #25 is shown in the next slide image.

We will introduce a vertical line between the third column and the fourth column to show the augmented matrix. The next step is to write the system of linear equations, we have three rows, which means that we will write three systems of equations that Can be written as follows: A- assume that we have a vertical column that has three rows with one column that is (X Y Z)

The first row (1 -2 4) is to be multiplied by the column vector (X Y Z). The first system of linear equations is

X-2y+4Z=3.

The second row (0 1 2) is to be multiplied by the column vector (X Y Z). The second system of linear equation is 0X+Y+2Z=7.

The third row (0 0 1) is to be multiplied by the column vector (X Y Z). The third system of linear equations is 0X+0Y+1Z=2. The triangular form is shown for the three systems of equations in the next slide image.

Use back substitution to get the values of X& Y and Z. We start with the last row for which we have a Z value equal to 2. We move **back **to the second system of linear equations and substitute the value of Z in that equation to get the value of Y which is equal to 7-4=3.

We move **back **to the first system of linear equations and substitute the value of Y&Z in that equation to get the value of X which is equal to 3-8+6=1. It is important to check the Values of X& Y and Z in any of the three systems of equations to make sure that the solution is accurate.

### Second practice problem for back substitution.

1-For the second practice problem #27 for back substitution. A matrix is given in the row-echelon form. a) write the system of equations for which the given matrix is the augmented matrix. b) use back substitution to solve the system. The detail of the given matrix for practice problem #27 is shown in the next slide image.

We will introduce a vertical line between the fourth column and the fifth column to show the augmented matrix. The next step is to write the system of linear equations, we have four rows, which means that we will write four systems of equations that Can be written as follows: A- assume that we have a vertical column that has three rows with one column that is (X Y Z W)

The first row (11 2 3 -1) is to be multiplied by the column vector (X Y Z W). The first system of linear equations is

X+2Y+3Z-W=7.

The second row (0 1 -2 0) is to be multiplied by the column vector (X Y Z W). The second system of linear equation is 0X+Y-2Z+0W=5.

The third row (0 0 1 2) is to be multiplied by the column vector (X Y Z W). The third system of linear equation is 0X+0Y+1Z+2W=5.

The fourth row (0 0 0 1) is to be multiplied by the column vector (X Y Z W). The fourth system of linear equation is 0X+0Y+0Z+1W=3. Two systems of linear equations are shown in the next slide image.

Use back substitution to get the values of X& Y &Z and W. We start with the last row for which we have a W value equal to 3.

We move **back **to the third system of linear equations and substitute the value of W in that equation to get the value of Z which is equal to 5-6=-1. These are the arrangements for the 3×4 matrix to have a row echelon form.

We move **back **to the second system of linear equations and substitute the values of Z&W in that equation to get the value of Y which is equal to 3.

Finally, we move **back **to the first system of linear equations and substitute the values of Y & Z and W in that equation to get the value of X which is equal to 7.

It is important to check the Values of X& Y and Z in any of the three systems of equations to make sure that the solution is accurate.

In the next post, we will solve practice problems for Gauss Jordan elimination.

This is a link to the matrix calculator.

For a useful external link, math is fun for the matrix part.