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. $$\left\\{\begin{aligned} -x+y-z &=4 \\ x-y+2 z &=3 \\ 2 x-2 y+4 z &=6 \end{aligned}\right.$$

Short answer

Infinitely many solutions.

Step-by-step solution

- Write the system of equations in standard form

The given system of equations is already in standard form:\[-x + y - z = 4\]\[x - y + 2z = 3\]\[2x - 2y + 4z = 6\]

- Write the augmented matrix

Create the augmented matrix from the system:\[\begin{bmatrix} -1 & 1 & -1 & | & 4 \ 1 & -1 & 2 & | & 3 \ 2 & -2 & 4 & | & 6 \end{bmatrix}\]

- Perform row operations to row reduce the matrix

1. Swap Row 1 and Row 2 to make the leading coefficient of the first row 1:\[\begin{bmatrix} 1 & -1 & 2 & | & 3 \ -1 & 1 & -1 & | & 4 \ 2 & -2 & 4 & | & 6 \end{bmatrix}\]2. Add Row 1 to Row 2:\[\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 0 & 1 & | & 7 \ 2 & -2 & 4 & | & 6 \end{bmatrix}\]3. Subtract 2 times Row 1 from Row 3:\[\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 0 & 1 & | & 7 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]

- Interpret the row-reduced matrix

The row-reduced matrix is:\[\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 0 & 1 & | & 7 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]This shows that there are infinitely many solutions because there is at least one row of all zeros in the coefficient matrix, indicating that the system is dependent.

- Verify the result using a computer

Using a computer or a calculator, the row-reduced form confirms that the system has infinitely many solutions.

Key concepts

Linear Systems

A linear system is a collection of one or more linear equations involving the same set of variables. For example, the system given in the exercise involves three equations with three variables (x, y, z). The general form of these equations is:
\[-x + y - z = 4\],
\[x - y + 2z = 3\],
and \[2x - 2y + 4z = 6\].
These equations are already in 'standard form,' where each term is either a variable or a constant.

Solving a linear system means finding the values of the variables that satisfy all equations at the same time. These variables can have

• exactly one solution,
• no solutions,
• or infinitely many solutions.

The method used in this context to find the solution is row reduction.

Row Reduction

Row reduction is a key technique in linear algebra used to simplify a system of linear equations. This process involves performing a series of operations on the augmented matrix (constructed from the given system of equations) to achieve a simpler form, known as Row Echelon Form (REF) or Reduced Row Echelon Form (RREF).
Here’s a step-by-step outline to understand row reduction:

• Step 1 - Swap Rows: Changing the order of the equations to make calculations easier.
• Step 2 - Scaling Rows: Multiplying or dividing a row by a non-zero constant to get leading coefficients of 1.
• Step 3 - Row Addition or Subtraction: Adding or subtracting multiples of rows to eliminate variables and achieve the desired form. In our example, these steps transformed the matrix to:

\[\begin{bmatrix} 1 & -1 & 2 & | & 3 \ -1 & 1 & -1 & | & 4 \ 2 & -2 & 4 & | & 6 \ \end{bmatrix}\]
This step-by-step process is fundamental to solving linear systems efficiently.

Infinitely Many Solutions

When you row reduce a system of equations, the form and values of the result can indicate different types of solutions. An important scenario is when the system has infinitely many solutions. This happens when you have at least one row in the reduced matrix where all the coefficient entries are zero, but the augmented part (right-hand side) does not contradict this.

In our example, the final row-reduced matrix looks like this:
\[\begin{bmatrix} 1 & -1 & 2 & | & 3 \ 0 & 0 & 1 & | & 7 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]

The third row is \[0 \ 0 \ 0 \ | \ 0\], indicating that there’s no unique solution. Instead, we have a dependent system, allowing for infinitely many solutions. We say it’s dependent since the rows aren't all linearly independent. This implies the existence of at least one free variable.

Dependent Systems

A dependent system of linear equations occurs when the equations in the system are not all independent. In simpler terms, some of the equations are essentially 'multiples' of others, or can be derived from combining others.

In the exercise, after performing row reduction, the final matrix showed us that there was a row of zeros, meaning one of the equations does not provide new information. This confirms that the system has dependent equations.
When systems are dependent, they do not give you a unique solution but instead, many solutions that form a linear space. To express this, one or more of the variables can be assigned an arbitrary parameter while solving.
This means there is an infinite number of vectors that are solutions to the system, derived from adjusting the free variable(s). Understanding dependent systems is crucial in linear algebra and many application fields such as engineering and physics.