Question

Show that the following system has more than one solution. $$ \begin{aligned} 3 \mathrm{x}-\mathrm{y}+7 \mathrm{z} &=0 \\ 2 \mathrm{x}-\mathrm{y}+4 \mathrm{z} &=1 / 2 \\ \mathrm{x}-\mathrm{y}+\mathrm{z} &=1 \\ 6 \mathrm{x}-4 \mathrm{y}+10 \mathrm{z} &=3 \end{aligned} $$

Short answer

The given system of equations is inconsistent, as seen by the last row in the row-echelon form of its augmented matrix, which implies 0 = -1. Therefore, the system has no solution, and it is not possible for this system to have more than one solution.

Step-by-step solution

Write the Augmented Matrix

Write the given system of equations in the form of an augmented matrix: \[ \left[\begin{array}{ccc|c} 3 & -1 & 7 & 0 \\ 2 & -1 & 4 & 1/2 \\ 1 & -1 & 1 & 1 \\ 6 & -4 & 10 & 3 \end{array}\right] \]

Perform Gaussian Elimination

Perform row operations to get the matrix in row echelon form: First, swap row 1 with row 3: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 2 & -1 & 4 & 1/2 \\ 3 & -1 & 7 & 0 \\ 6 & -4 & 10 & 3 \end{array}\right] \] Next, use row 1 to eliminate x from other equations: Subtract 2 times row 1 from row 2: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & 2 & -1 \\ 3 & -1 & 7 & 0 \\ 6 & -4 & 10 & 3 \end{array}\right] \] Subtract 3 times row 1 from row 3: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & 2 & -1 \\ 0 & 2 & 4 & -3 \\ 6 & -4 & 10 & 3 \end{array}\right] \] Subtract 6 times of row 1 from row 4: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & 2 & -1 \\ 0 & 2 & 4 & -3 \\ 0 & 2 & 4 & -3 \end{array}\right] \] Now, use row 2 to eliminate y from other equations: Subtract 2 times row 2 from row 3, and subtract row 2 from row 4: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & -1 \\ 0 & 1 & 2 & -1 \end{array}\right] \] Now, use row 3 to eliminate z from other equations: Add row 3 to row 1, and subtract row 3 from row 4: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 0 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & -1 \\ 0 & 1 & 2 & -2 \end{array}\right] \] Continue with the last row operation: Add row 4 to row 3: \[ \left[\begin{array}{ccc|c} 1 & -1 & 1 & 0 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right] \]

Analyze the Row-Echelon Form

We can see that there are more equations than unknowns and the last row is inconsistent (implies 0 = -1). This system has no solution. Thus, it is not possible for this system to have more than one solution.

Key concepts

Gaussian Elimination

Gaussian Elimination is a method to systematically solve systems of linear equations. It consists of performing a sequence of operations on the augmented matrix of the system to transform it into a simpler form, ideally into what is known as row echelon form. The operations include swapping two rows, multiplying a row by a nonzero scalar, and adding or subtracting multiples of one row to another row.

These operations are designed to achieve one main goal: to simplify the system so that each row represents a single equation, and each column lines up with a specific variable. The process simplifies solving the equations mainly by eliminating variables, step by step. When the system has been transformed into the row echelon form, it can be easily solved by back substitution starting from the last equation.

Augmented Matrix

The augmented matrix of a system of linear equations is essentially a compact way to write down the system, combining the coefficients of the variables and the constants from the equations' right-hand sides into a single matrix. In such a matrix, the vertical bar separates the coefficients on the left from the constants on the right.

The advantage of the augmented matrix is that it allows us to apply row operations to both the coefficients and the constants simultaneously. By doing so, this maintains the integrity of the equations and helps us move towards a solution. The augmented matrix is the starting point for Gaussian Elimination and is crucial for understanding the structure and solutions of a system of linear equations.

Row Echelon Form

Row echelon form is a stage in the Gaussian Elimination process where the augmented matrix has been transformed to satisfy specific conditions. Every nonzero row has a leading coefficient (also known as the pivot) which is to the right of the leading coefficient of the row above it. Also, all rows consisting entirely of zeros (if there are any) are at the bottom of the matrix.

Having a matrix in row echelon form is useful because it corresponds to a system of equations that are easier to solve. Each equation should have one more variable than the previous, lining up directly above or below its respective leading coefficient, which makes back substitution possible. Sometimes, during Gaussian Elimination, inconsistencies can be revealed in this form, indicating the system has no solution, or it has infinitely many solutions depending on the situation.

Inconsistent System

An inconsistent system is one in which no solution exists that satisfies all equations simultaneously. This usually becomes apparent when transforming the system into row echelon form during Gaussian Elimination.

For example, if a row has zeros in all columns corresponding to variables but has a nonzero constant in the augmented part (as in the last step of the provided solution, \[0 & 0 & 0 & -1\]), this indicates that the system requires 0 to be equal to a nonzero number, which is impossible. This conflict means there is no set of values for the variables that can satisfy all the original equations, hence the system is inconsistent.