Problem 36 - College Algebra with Modeling and Visualization

Question

Use Gaussian elimination with backward substitution to solve the system of linear equations. Write the solution as an ordered pair or an ordered triple whenever possible. $$ \begin{array}{r} x+y+z=3 \\ x+y+2 z=4 \\ 2 x+2 y+3 z=7 \end{array} $$

Step-by-Step Solution

Answer

The system has infinitely many solutions given by \((x, y, z) = (2 - s, s, 1)\) for any real \( s \).

Step 1: Write the system in augmented matrix form

First, we need to convert the given linear system into its augmented matrix form. The coefficients of the system \( x + y + z = 3 \), \( x + y + 2z = 4 \), and \( 2x + 2y + 3z = 7 \) are used to form the matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \1 & 1 & 2 & | & 4 \2 & 2 & 3 & | & 7 \\end{bmatrix}\]

Step 2: Perform Row Operations to Get Upper Triangular Form

We aim to create zeros below the first pivot (1st row, 1st column).- Subtract the 1st row from the 2nd row: \( R_2 - R_1 \rightarrow R_2 \).- Subtract 2 times the 1st row from the 3rd row: \( R_3 - 2R_1 \rightarrow R_3 \).The new matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & 0 & 1 & | & 1 \0 & 0 & 1 & | & 1 \\end{bmatrix}\]

Step 3: Check for Inconsistencies

Notice the similarity between the 2nd and 3rd rows. To see if this system is consistent, subtract the 2nd row from the 3rd row: \( R_3 - R_2 \rightarrow R_3 \).The matrix simplifies to:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & 0 & 1 & | & 1 \0 & 0 & 0 & | & 0 \\end{bmatrix}\]The third row, \( 0 = 0 \), indicates a free variable.

Step 4: Express in Terms of Free Variable

Since the third row is now zero, the system has a free variable, say \( z = t \). The second row gives \( z = 1 \). One could choose \( z \) as a leading variable too. Suppose \( z = 1 \):- 2nd row implies \( z = 1 \).- Substitute \( z = 1 \) into original first equation: \( x + y + 1 = 3 \rightarrow x + y = 2 \).

Step 5: Backward Substitution

To find the values of \( x \) and \( y \), assign \( y = s \) (as a free variable).- From \( x + y = 2 \), get \( x = 2 - s \).- Since \( z = 1 \), the solution is:\((x, y, z) = (2 - s, s, 1)\) where \( s \in \mathbb{R} \).

Step 6: Conclusion

The system has infinitely many solutions parametrized by the free variable \( s \) with the forms \((x, y, z) = (2 - s, s, 1)\), where \( s \) can be any real number.

Key Concepts

Linear Equations in Gaussian Elimination

Linear equations are the foundation of the system we are trying to solve using Gaussian elimination. In this context, linear equations involve variables like \(x\), \(y\), and \(z\), each appearing to the first power. The exercise presents the system of linear equations as follows:

• \(x + y + z = 3\)
• \(x + y + 2z = 4\)
• \(2x + 2y + 3z = 7\)

Using these equations effectively involves understanding their coefficients and how these numbers play a crucial role in the Gaussian elimination process. The main goal is to manipulate these equations to find the values of the variables that satisfy all equations simultaneously.

Understanding the Augmented Matrix

The augmented matrix is a vital tool in solving systems of linear equations using Gaussian elimination. It combines both the coefficients from the linear equations and their constants into a single matrix format. For our exercise, the original system of equations is transformed into the following augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \1 & 1 & 2 & | & 4 \2 & 2 & 3 & | & 7 \\end{bmatrix}\]This matrix displays each equation's coefficients as rows, with a vertical bar separating them from the constants on the right. Row operations are then performed to simplify this matrix into an upper triangular form, which makes it easier to deduce the values of the variables through subsequent steps.

Employing Backward Substitution

Backward substitution comes into play after transforming the augmented matrix into an upper triangular form. This structured form has non-zero elements at the top-left and zeros below. In our exercise, performing row operations leads to the form:\[\begin{bmatrix}1 & 1 & 1 & | & 3 \0 & 0 & 1 & | & 1 \0 & 0 & 0 & | & 0 \\end{bmatrix}\]Backward substitution involves starting with the bottom-most non-zero row to find the first variable. Here, that means beginning with \(z = 1\). We then substitute this value into the above rows to find other variables. By applying backward substitution, we derive the full solution, determining relationships among the variables considering any free variables present.

The Role of Free Variables

Free variables emerge when there are fewer equations than unknowns, or when the system provides an infinite number of solutions. In our exercise, after simplifying the matrix, we are left with the third row showing \(0 = 0\), which indicates a free variable is present.Assume \(z = t\), where \(t\) is a free variable representing any real number. Consequently, this gives rise to infinitely many solutions where \(z\) can be any value. We express the solution set as:

• For \(z = 1\), substitute into \(x + y + z = 3\), leading to \(x + y = 2\).
• Given a new free variable \(y = s\), we get \(x = 2 - s\).

Ultimately, the solution set is written as \((x, y, z) = (2 - s, s, 1)\), where \(s\) is any real number, illustrating the multiple possibilities for \(y\) while satisfying the entire system.