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. $$\left\\{\begin{array}{l} 2 x+y-z=2 \\ 4 x+y-2 z=3 \end{array}\right.$$

Short answer

The system has an infinite number of solutions.

Step-by-step solution

– Write the system in augmented matrix form

Convert the given system of equations into an augmented matrix. The system is: 1. \(2x + y - z = 2\)2. \(4x + y - 2z = 3\)The augmented matrix for this system is:\[ \begin{bmatrix} 2 & 1 & -1 & | & 2 \ 4 & 1 & -2 & | & 3 \end{bmatrix} \]

– Apply row operations to obtain row echelon form

We want to simplify the matrix to upper triangular form. Start with the matrix from Step 1:\[ \begin{bmatrix} 2 & 1 & -1 & | & 2 \ 4 & 1 & -2 & | & 3 \end{bmatrix} \]- Divide the first row by 2 to normalize the leading coefficient:\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 4 & 1 & -2 & | & 3 \end{bmatrix} \]- Subtract 4 times the first row from the second row to eliminate the leading coefficient in the second row:\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 0 & -1 & 0 & | & -1 \end{bmatrix} \]

– Simplify the second row

Multiply the entire second row by -1 to normalize, resulting in:\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 0 & 1 & 0 & | & 1 \end{bmatrix} \]

– Back substitute to determine the solution

Use back substitution to solve for the variables.From the second row, we have: \(y = 1\)Substitute \(y = 1\) into the first row:\( x + \frac{1}{2}(1) - \frac{1}{2}z = 1 \Rightarrow x + \frac{1}{2} - \frac{1}{2}z = 1 \Rightarrow x - \frac{1}{2}z = \frac{1}{2}\)Assume \(z = 0\), giving:\( x = \frac{1}{2}\)Thus, \(z\) is a free variable, leading to infinitely many solutions.

Key concepts

Augmented Matrix

When dealing with systems of linear equations, an augmented matrix is a convenient way to represent the system. The augmented matrix combines the coefficients of the variables and the constants from the right side of the equations into one matrix.
For example, the system of equations:
\[ \begin{cases} 2x + y - z = 2 \ 4x + y - 2z = 3 \ \end{cases} \]
can be written as an augmented matrix:
\[ \begin{bmatrix} 2 & 1 & -1 & | & 2 \ 4 & 1 & -2 & | & 3 \ \end{bmatrix} \]
Here, each row represents an equation, and each column represents the coefficients of a variable or the constants on the right-hand side. The vertical bar separates the coefficients from the constants, offering a clear visual distinction.

Row Operations

Row operations are essential tools used to simplify augmented matrices. These operations allow us to manipulate the matrix to achieve row-echelon or reduced row-echelon form, making it easier to solve the system of equations. There are three main types of row operations:

• Row Swapping: Exchange two rows.
• Row Multiplication: Multiply a row by a non-zero scalar.
• Row Addition: Add or subtract a multiple of one row to another row.

For instance, in our example:

• Divide the first row by 2 to normalize it:
  \[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 4 & 1 & -2 & | & 3 \ \end{bmatrix} \]
• Subtract 4 times the first row from the second row:
  \[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 0 & -1 & 0 & | & -1 \ \end{bmatrix} \]
• Multiply the second row by -1:
  \[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 0 & 1 & 0 & | & 1 \ \end{bmatrix} \]

These operations help in systematically reducing the matrix.

Back Substitution

Once the augmented matrix is in row-echelon form, we can use back substitution to find the solutions to the system of equations. Back substitution involves solving the equations starting from the last row and moving upwards. Each solved variable is substituted back into the previous rows.
In our example, the final row-echelon form is:
\[ \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} & | & 1 \ 0 & 1 & 0 & | & 1 \ \end{bmatrix} \]
Start with the second row:
\[ y = 1 \]
Then back-substitute this value into the first row:
\[ x + \frac{1}{2}(1) - \frac{1}{2}z = 1 \]
Simplify to find:
\[ x - \frac{1}{2}z = \frac{1}{2} \]
Here, we see that the variable z is free, meaning we can assign it any value.

Infinite Solutions

A system of equations has infinite solutions when there are free variables present. Free variables occur when the matrix has rows that are not fully independent of each other, leading to a situation where we can assign arbitrary values to some variables.
In our example, the variable \(z\) is free:
\[ z = 0 \]
Substituting \(z = 0\) into:
\[ x - \frac{1}{2}(0) = \frac{1}{2} \]
Gives:
\[ x = \frac{1}{2} \]
But we could have chosen any value of \(z\), leading to different values for \(x\).
Hence, the system has infinitely many solutions because we can vary \(z\) freely and find corresponding \(x\) and \(y\) that satisfy the system.