Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{r} x+2 y-z=-3 \\ 2 x-4 y+z=-7 \\ -2 x+2 y-3 z=4 \end{array}\right. $$

Short answer

The solution is x = 0, y = 1, and z = 7.

Step-by-step solution

- Write the augmented matrix

First, represent the given system of equations as an augmented matrix. o \[ \begin{pmatrix} 1 & 2 & -1 & | & -3 \ 2 & -4 & 1 & | & -7 \ -2 & 2 & -3 & | & 4 \ \end{pmatrix} \]

- Make the first element of the first row 1

The first element of the first row is already 1, so we don't need to make any changes.

- Eliminate the first element of the second and third rows

Subtract 2 times the first row from the second row and add 2 times the first row to the third row. \[ \text{New Row 2} = \text{Row 2} - 2 \times \text{Row 1} \ \text{New Row 3} = \text{Row 3} + 2 \times \text{Row 1} \] New matrix: \[ \begin{pmatrix} 1 & 2 & -1 & | & -3 \ 0 & -8 & 3 & | & -1 \ 0 & 6 & -5 & | & -2 \end{pmatrix} \]

- Make the second element of the second row 1

Divide the second row by -8. \[ \text{New Row 2} = \frac{\text{Row 2}}{-8} \]New matrix:\[ \begin{pmatrix} 1 & 2 & -1 & | & -3 \ 0 & 1 & -\frac{3}{8} & | & \frac{1}{8} \ 0 & 6 & -5 & | & -2 \end{pmatrix} \]

- Eliminate the second element of the first and third rows

Subtract 2 times the second row from the first row and subtract 6 times the second row from the third row.\[ \text{New Row 1} = \text{Row 1} - 2 \times \text{Row 2} \ \text{New Row 3} = \text{Row 3} - 6 \times \text{Row 2} \] New matrix: \[ \begin{pmatrix} 1 & 0 & -\frac{1}{4} & | & -\frac{7}{4} \ 0 & 1 & -\frac{3}{8} & | & \frac{1}{8} \ 0 & 0 & -\frac{1}{4} & | & -\frac{7}{4} \end{pmatrix} \]

- Make the third element of the third row 1

Divide the third row by \frac{-1}{4}. \[ \text{New Row 3} = \frac{\text{Row 3}}{\frac{-1}{4}} \] New matrix: \[ \begin{pmatrix} 1 & 0 & -\frac{1}{4} & | & -\frac{7}{4} \ 0 & 1 & -\frac{3}{8} & | & \frac{1}{8} \ 0 & 0 & 1 & | & 7 \end{pmatrix} \]

- Eliminate the third element of the first and second rows

Add \frac{1}{4} times the third row to the first row and add \frac{3}{8} times the third row to the second row. \[ \text{New Row 1} = \text{Row 1} + \frac{1}{4} \times \text{Row 3} \ \text{New Row 2} = \text{Row 2} + \frac{3}{8} \times \text{Row 3} \] New matrix: \[ \begin{pmatrix} 1 & 0 & 0 & | & 0 \ 0 & 1 & 0 & | & 1 \ 0 & 0 & 1 & | & 7 \end{pmatrix} \]

- Write the solution

The matrix is now in reduced row echelon form. Thus, the solutions to the system are x = 0, y = 1, and z = 7.

Key concepts

Row Operations

Row operations are essential when solving systems of equations using matrices. They are actions you perform on the rows of a matrix to get it into a simpler form. There are three main types of row operations:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another row

Each row operation modifies the matrix while keeping it equivalent to the original system of equations. This means the solutions to the system do not change, but the matrix becomes easier to work with.
In our example, we started by writing the equations as an augmented matrix and used these operations to simplify the matrix.

Augmented Matrix

An augmented matrix is a way to represent a system of linear equations. It combines the coefficients of the variables and the constants of the equations into one matrix. For instance, the system
\[ \begin{array}{r} x+2y-z=-3 \ 2x-4y+z=-7 \ -2x+2y-3z=4 \ \end{array} \]
is converted to the augmented matrix:
\[ \begin{pmatrix} 1 & 2 & -1 & | & -3 \ 2 & -4 & 1 & | & -7 \ -2 & 2 & -3 & | & 4 \ \end{pmatrix} \]
Here, each row represents an equation, and the vertical line separates the coefficients and the constants. This form simplifies the application of row operations and helps in visualizing the steps needed to solve the system.

Reduced Row Echelon Form

Reduced Row Echelon Form (RREF) is the simplified version of a matrix that makes solving systems of equations straightforward. A matrix is in RREF if:

• All leading entries are 1
• Each leading 1 is the only non-zero entry in its column
• The leading 1 in any row is to the right of the leading 1 in the row above
• Rows with all zeroes are at the bottom

For example, the matrix
\[ \begin{pmatrix} 1 & 0 & 0 & | & 0 \ 0 & 1 & 0 & | & 1 \ 0 & 0 & 1 & | & 7 \ \end{pmatrix} \]
is in RREF. The solutions can be read directly from this form. In our example, after applying row operations, we ended up with a matrix in RREF from which we deduced the solutions: x = 0, y = 1, z = 7.

Inconsistent Systems

An inconsistent system of equations has no solution. This occurs when the equations represent parallel lines that never intersect or otherwise form a contradictory system.
In matrix terms, this is indicated by a row where all entries except the constant on the right are zero, but the constant is non-zero, like:
\[ \begin{pmatrix} 0 & 0 & 0 & | & 1 \ \end{pmatrix} \]
This tells us that the equations conflict with each other. In our example, the system was consistent, as we did not encounter such a row. Therefore, we could find a unique solution.