Problem 13 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{aligned} &3 x+y+z=0\\\ &4 x+2 y+z=1\\\ &2 x-2 y-z=2 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( (x, y, z) = \left( \frac{1}{2}, \frac{1}{2}, -2 \right) \).

Step 1: Set Up the Augmented Matrix

First, we represent the system of equations as an augmented matrix: \[\begin{bmatrix} 3 & 1 & 1 & | & 0 \ 4 & 2 & 1 & | & 1 \ 2 & -2 & -1 & | & 2 \end{bmatrix}\]

Step 2: Perform Row Operations to Achieve Row Echelon Form

We first eliminate the element below the first pivot (3) in the second row by substituting Row 2 with the result of \(\text{Row}_2 - \frac{4}{3}\text{Row}_1\). Similarly, eliminate below Row 1 column 1 for Row 3 by replacing \(\text{Row}_3\) with \(\text{Row}_3 - \frac{2}{3}\text{Row}_1\). This gives:\[\begin{bmatrix} 3 & 1 & 1 & | & 0 \ 0 & \frac{2}{3} & \frac{-1}{3} & | & 1 \ 0 & -\frac{8}{3} & -\frac{5}{3} & | & 2 \end{bmatrix}\]

Step 3: Perform Additional Row Operations

Next, eliminate the coefficient in Row 3 column 2. Replace Row 3 with \( \text{Row}_3 + 4\text{Row}_2 \):\[\begin{bmatrix} 3 & 1 & 1 & | & 0 \ 0 & \frac{2}{3} & \frac{-1}{3} & | & 1 \ 0 & 0 & -3 & | & 6 \end{bmatrix}\]

Step 4: Back Substitution

Solve the equations from the bottom upwards. Starting with Row 3, solve for \(z\) from \(-3z = 6\):\[ z = -2 \]Then substitute \(z = -2\) into Row 2 to solve for \(y\):\[ \frac{2}{3}y - \frac{1}{3}(-2) = 1 \]\[ \frac{2}{3}y + \frac{2}{3} = 1 \]\[ \frac{2}{3}y = \frac{1}{3} \]\[ y = \frac{1}{2} \]Finally, substitute \(y = \frac{1}{2}\) and \(z = -2\) into Row 1 to solve for \(x\):\[ 3x + \frac{1}{2} - 2 = 0 \]\[ 3x - \frac{3}{2} = 0 \]\[ 3x = \frac{3}{2} \]\[ x = \frac{1}{2} \]

Step 5: Write the Solution

The solution to the system of equations is \((x, y, z) = \left( \frac{1}{2}, \frac{1}{2}, -2 \right)\).

Key Concepts

Augmented Matrix

To transform a system of linear equations into something more manageable, an augmented matrix is a great tool. It basically converts the entire system into a matrix form, which is easier to work with, especially for visualizing row operations. In the provided exercise, the system of equations is:

• 3x + y + z = 0
• 4x + 2y + z = 1
• 2x - 2y - z = 2

Representing the above equations as an augmented matrix, you get:divide the matrix into two parts with a vertical line after the coefficient of variables, forming:\[\begin{bmatrix}3 & 1 & 1 & | & 0 \4 & 2 & 1 & | & 1 \2 & -2 & -1 & | & 2\end{bmatrix}\] Here, the numbers on the left are the coefficients of your variables framed in a matrix, while the numbers on the right of the vertical line are your constants. This matrix serves as the first step in applying further techniques to solve the system under the method of Gaussian elimination.

Row Operations

Row operations are key in transforming the augmented matrix into a more useful form. They involve three main types:

• Swapping two rows.
• Multiplying a row by a nonzero scalar.
• Replacing a row by the sum of it and another row multiplied by a scalar.

In the exercise, the aim of row operations is to achieve the row-echelon form, where you can see leading entries (pivots) forming a staircase pattern down the matrix. This is done because it's much easier to work with a system in this form. The major operations here include:

• Modifying Row 2 by replacing it with \( \text{Row}_2 - \frac{4}{3} \text{Row}_1 \), effectively eliminating the x-term from Row 2.
• Changing Row 3 to \( \text{Row}_3 - \frac{2}{3} \text{Row}_1 \) to remove the x-term from Row 3.
• Further simplifying Row 3 by performing \( \text{Row}_3 + 4 \text{Row}_2 \) to clean out the y-term.

Post row operations, the matrix appears as:\[\begin{bmatrix}3 & 1 & 1 & | & 0 \0 & \frac{2}{3} & \frac{-1}{3} & | & 1 \0 & 0 & -3 & | & 6\end{bmatrix}\] These manipulations define the core of Gaussian elimination, setting the stage for back substitution.

Back Substitution

After performing row operations and reaching an echelon form, back substitution steps in to finally solve the system. Starting from the last row where typically most operations are done, solve for the remaining unknowns one by one.In the given matrix: \[\begin{bmatrix}3 & 1 & 1 & | & 0 \0 & \frac{2}{3} & \frac{-1}{3} & | & 1 \0 & 0 & -3 & | & 6\end{bmatrix}\]Start with the last equation: \(-3z = 6\), immediately providing \(z = -2\). Plug this back into the preceding rows to solve for \(y\) and \(x\).- For the second row, \(\frac{2}{3}y - \frac{1}{3}(-2) = 1\), simplify to solve \(y = \frac{1}{2}\).- Finally, substitute the found values of \(y\) and \(z\) into the first row equation, \(3x + \frac{1}{2} - 2 = 0\), to find \(x = \frac{1}{2}\).Back substitution uses the upper triangular form achieved from row operations to easily determine each variable, leading to the system's solution: \((x, y, z) = \left(\frac{1}{2}, \frac{1}{2}, -2 \right)\). This method ensures that, once the matrix is in echelon form, each subsequent variable is easily resolved one after the other.