Question

For each of the following problems write and row reduce the augmented matrix to find out whether the given set of equations has exactly one solution, no solutions, or an infinite set of solutions. Check your results by computer. Warning hint: Be sure your equations are written in standard form. Comment: Remember that the point of doing these problems is not just to get an answer (which your computer will give you), but to become familiar with the terminology, ideas, and notation we are using. $$\left\\{\begin{aligned} x+2 y-z &=1 \\ 2 x+3 y-2 z &=-1 \\ 3 x+4 y-3 z &=-4 \end{aligned}\right.$$

Short answer

The system has no solution.

Step-by-step solution

Write the system of equations in matrix form

First, express the given linear equations as an augmented matrix. The system of equations is given by: \[ \begin{cases} x + 2y - z = 1 \ 2x + 3y - 2z = -1 \ 3x + 4y - 3z = -4 \end{cases} \] The corresponding augmented matrix is: \[ \begin{pmatrix} 1 & 2 & -1 & \, | \, & 1 \ 2 & 3 & -2 & \, | \, & -1 \ 3 & 4 & -3 & \, | \, & -4 \end{pmatrix} \]

Transform the augmented matrix to Row Echelon Form (REF)

Use row operations to transform the matrix into Row Echelon Form. We start with the matrix: \[ \begin{pmatrix} 1 & 2 & -1 & \, | \, & 1 \ 2 & 3 & -2 & \, | \, & -1 \ 3 & 4 & -3 & \, | \, & -4 \end{pmatrix} \] First, subtract 2 times the first row from the second row: \[ R2 = R2 - 2R1 \rightarrow \begin{pmatrix} 1 & 2 & -1 & \, | \, & 1 \ 0 & -1 & 0 & \, | \, & -3 \ 3 & 4 & -3 & \, | \, & -4 \end{pmatrix} \] Then, subtract 3 times the first row from the third row: \[ R3 = R3 - 3R1 \rightarrow \begin{pmatrix} 1 & 2 & -1 & \, | \, & 1 \ 0 & -1 & 0 & \, | \, & -3 \ 0 & -2 & 0 & \, | \, & -7 \end{pmatrix} \]

Continue reducing to REF

Next, subtract 2 times the second row from the third row: \[ R3 = R3 - 2R2 \rightarrow \begin{pmatrix} 1 & 2 & -1 & \, | \, & 1 \ 0 & -1 & 0 & \, | \, & -3 \ 0 & 0 & 0 & \, | \, & -1 \end{pmatrix} \]

Interpret the Row Echelon Form

In REF, the last row indicates \( 0 = -1 \), which is a contradiction. Therefore, the system of equations has no solution.

Key concepts

augmented matrix

An augmented matrix is a helpful way to represent a system of linear equations. It combines the coefficient matrix with the constants from the equations into a single matrix.
For example, consider the system of equations: \[ \begin{cases} x + 2y - z = 1 \ 2x + 3y - 2z = -1 \ 3x + 4y - 3z = -4 \ \ \right.\] To write these in augmented matrix form, follow the steps below:
Write down the coefficients of the variables in a matrix. Then, draw a vertical line to separate the coefficients from the constants on the right-hand side.
This yields: \[ \begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ 2 & 3 & -2 & \, | \ & -1 \ 3 & 4 & -3 & \, | \ & -4 \ \right.\]
This format makes it easy to apply matrix operations and maintain the same logical structure as the original equations.

row reduction

Row reduction, or Gaussian elimination, is a series of operations performed on an augmented matrix to solve a system of linear equations. The goal is to simplify the matrix to a form where the solution can be easily identified.
The operations you can use include:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting a multiple of one row to another row

For our example, we begin with: \[\begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ 2 & 3 & -2 & \, | \ & -1 \ 3 & 4 & -3 & \, | \ & -4 \right.\]
First, subtract 2 times the first row from the second row: \[R2 = R2 - 2R1 \rightarrow \begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ 0 & -1 & 0 & \, | \ & -3 \ 3 & 4 & -3 & \, | \ & -4 \right.\]
Then, subtract 3 times the first row from the third row: \[ R3 = R3 - 3R1 \rightarrow \begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ 0 & -1 & 0 & \, | \ & -3 \ 0 & -2 & 0 & \, | \ & -7 \right.\]
Row reduction simplifies the matrix step by step, leading us closer to the solution.

row echelon form

Row echelon form (REF) is a simplified version of a matrix achieved through row reduction. In REF:

• All zero rows (rows with all zero entries) are at the bottom of the matrix
• The leading entry (first non-zero number from the left) of each non-zero row after the first occurs to the right of the leading entry of the previous row
• The leading entry in any non-zero row is 1, and this 1 is the only non-zero entry in its column

Row reduction of our matrix: \[\begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ 0 & -1 & 0 & \, | \ & -3 \ 0 & -2 & 0 & \, | \ & -7 \right.\]
Subtract 2 times the second row from the third row to get: \[R3 = R3 - 2R2 \rightarrow \begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ 0 & -1 & 0 & \, | \ & -3 \ 0 & 0 & 0 & \, | \ & -1 \right.\]
This specialized form helps in determining whether a system has a solution and identifying that solution if it exists.

no solution

At times, row reduction of an augmented matrix reveals that a system of equations has no solution. This occurs when you end up with a row that implies a contradiction, like:
\[\begin{pmatrix} 1 & 2 & -1 & \, | \ & 1 \ \ \0 & -1 & 0 & \, | \ & -3 \ \0 & 0 & 0 & \, | \ & -1 \right.\]
The third row now reads \[ 0 = -1 \], which is clearly impossible. Therefore, this system of equations has no solution.
Whenever the row echelon form includes such a row, you have an inconsistent system. This means that no combination of variables will satisfy all the given equations at once, confirming that there is no solution.