Problem 33 - 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. $$ \begin{aligned} 3 x+y+3 z &=14 \\ x+y+z &=6 \\ -2 x-2 y+3 z &=-7 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \((5, 2, -1)\).

Step 1: Write the augmented matrix

For the given system of equations, first, create an augmented matrix to represent it. The system is: \[\begin{align*} 3x + y + 3z &= 14 \ x + y + z &= 6 \-2x - 2y + 3z &=-7 \end{align*} \] The augmented matrix is: \[\begin{bmatrix}3 & 1 & 3 & | & 14 \1 & 1 & 1 & | & 6 \-2 & -2 & 3 & | & -7 \end{bmatrix}\]

Step 2: Perform row operations to obtain an upper triangular matrix

Start by using the first row to eliminate the coefficients below the first pivot element. - Multiply row 2 by 3 and subtract from row 1: \[R_1' = R_1 - 3 imes R_2 ightarrow (3-3) = 0, (1-3) = -2, (3-3) = 0, (14-18) = -4 \]Resulting in: \[\begin{bmatrix}3 & 1 & 3 & | & 14 \0 & -2 & 0 & | & -4 \-2 & -2 & 3 & | & -7 \end{bmatrix}\]- Add twice the second row to the third row:\[R_3' = R_3 + 2 imes R_2 ightarrow (-2+0)=-2, (-2-4)=-6, (3+0)=3, (-7-8)=-15\]Resulting in:\[\begin{bmatrix}3 & 1 & 3 & | & 14 \0 & -2 & 0 & | & -4 \0 & -6 & 3 & | & -15 \end{bmatrix}\]

Step 3: Further elimination to simplify triangle

Divide row 2 by -2 to simplify:\[R_2' = \frac{1}{-2}R_2 = \begin{bmatrix}0 & 1 & 0 & | & 2 \end{bmatrix}\]Replace row 2:\[\begin{bmatrix}3 & 1 & 3 & | & 14 \0 & 1 & 0 & | & 2 \0 & -6 & 3 & | & -15 \end{bmatrix}\]Now, eliminate the second pivot row from the third row by multiplying the current second row by 6 and adding to the third:\[R_3' = R_3 + 6 imes R_2 ightarrow (0+0) = 0, (-6+6) = 0, (3+0) = 3, (-15+12) = -3 \]Resulting in:\[\begin{bmatrix}3 & 1 & 3 & | & 14 \0 & 1 & 0 & | & 2 \0 & 0 & 3 & | & -3 \end{bmatrix}\]

Step 4: Backward substitution to find solutions

Begin solving the system by setting each variable in terms of the matrix: From row 3: \[3z = -3 \rightarrow z = -1\]From row 2: \[y = 2\]Substitute these values into row 3 to solve for \(x\):\[3x + 1 \times 2 + 3 \times (-1) = 14 \rightarrow 3x + 2 - 3 = 14 \rightarrow 3x = 15 \rightarrow x = 5\]

Step 5: Write the final solution as an ordered triple

Combine the values \(x = 5\), \(y = 2\), and \(z = -1\) into an ordered triple representing the solution.

Key Concepts

Linear Equations

Linear equations involve variables raised only to the first power. They represent straight lines when graphed on a coordinate plane. In this exercise, we have a system of three linear equations in three variables, which makes it a 3x3 system. Each equation in this system can be imagined as a plane in three-dimensional space.

• The first equation is \(3x + y + 3z = 14\).
• The second equation is \(x + y + z = 6\).
• The third equation is \(-2x - 2y + 3z = -7\).

The main goal in solving such systems is to find the values of the variables that satisfy all equations simultaneously. These values correspond to the point where all planes intersect, representing the only solution of the system if it exists.

Augmented Matrix

To solve systems of linear equations efficiently, we use an augmented matrix. This matrix combines the coefficients and constants of the equations into a single rectangular array. Each row of the matrix represents an equation, and each column represents the coefficients of the variables. The final column represents the constants from each equation. For our system, the augmented matrix looks like this:\[\begin{bmatrix}3 & 1 & 3 & | & 14 \1 & 1 & 1 & | & 6 \-2 & -2 & 3 & | & -7 \end{bmatrix}\]The vertical line separates the coefficients from the constants. The next step in Gaussian elimination is to use row operations to simplify this matrix into an upper triangular form, making it easier to perform backward substitution.

Backward Substitution

Backward substitution is a method used after transforming the augmented matrix into an upper triangular form through Gaussian elimination. At this stage, the matrix is arranged such that all elements below the main diagonal are zeros. This simplifies solving for the variables.Starting with the last row, which corresponds to the simplest equation, you solve for its variable. Then, you substitute this value into the row above it to solve for the next variable. Continue this process until all variables are solved. In our exercise:- From the third row, \(3z = -3\) leads to \(z = -1\).- Row two becomes \(y = 2\) after substitution.- Substitute \(y\) and \(z\) values back into the first row, \(3x + 1 \times 2 + 3 \times (-1) = 14\), resulting in \(x = 5\).

Solution of System

The solution of a system of linear equations is the set of values for the variables that satisfy every equation in the system. For systems with three variables, this solution is often represented as an ordered triple.From the previous steps in Gaussian elimination and backward substitution, we determined that: - \(x = 5\)- \(y = 2\)- \(z = -1\)These values form the solution \((5, 2, -1)\). This triple indicates the intersection point of the three planes represented by the equations in three-dimensional space. Solving a system of linear equations provides meaningful insights across various disciplines, such as physics, economics, and engineering, where such equations model relationships between multiple factors.