Problem 31 - Precalculus: Mathematics for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$\left\\{\begin{aligned} x+y+z &=2 \\ 2 x-3 y+2 z &=4 \\ 4 x+y-3 z &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (1, 0, 1)\).

Step 1: Set up the augmented matrix

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

Step 2: Perform Gaussian elimination

Begin by eliminating the x-terms below the first row. Subtract 2 times the first row from the second row, and 4 times the first row from the third.\[\begin{bmatrix}1 & 1 & 1 & | & 2 \2-2(1) & -3-2(1) & 2-2(1) & | & 4-2(2) \4-4(1) & 1-4(1) & -3-4(1) & | & 1-4(2)\end{bmatrix}=\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & -3 & -7 & | & -7\end{bmatrix}\]

Step 3: Eliminate y-term from third row

Now, eliminate the y-term from the third row by adding 3/5 of the second row to the third row.\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & -3+3(-5)/5 & -7 & | & -7+0\end{bmatrix}=\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & 0 & -7 & | & -7\end{bmatrix}\]

Step 4: Back substitution

With the triangular form achieved, solve for each variable starting from the last row. \From the third row: \\(-7z = -7\), thus, \(z = 1\).\Substitute \(z = 1\) into the second row: \\( -5y = 0 \), therefore, \( y = 0 \). \Finally, substitute \(y = 0\) and \(z = 1\) into the first row:\(x + 0 + 1 = 2 \), resulting in \(x = 1\).

Step 5: Present the solution

The solution to the system of equations is \((x, y, z) = (1, 0, 1)\).

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple equations with the same set of unknown variables. In simpler words, you want to find the values of these variables that make all the equations true at the same time. For instance, in our original exercise, we have three equations:

• \(x+y+z=2\)
• \(2x-3y+2z=4\)
• \(4x+y-3z=1\)

Here, the unknown variables are \(x\), \(y\), and \(z\). The challenge is to solve these equations simultaneously. It means finding the values of \(x\), \(y\), and \(z\) that work in all these equations. Such systems can have a unique solution, infinitely many solutions, or no solution at all. Gaussian elimination is a great tool to solve systems with a unique solution.

Augmented Matrix

To apply techniques like Gaussian elimination efficiently, we often represent a system of equations as an augmented matrix. This matrix simplifies handling the equations.
Let's take our system and convert it into an augmented matrix:

• Equation coefficients form the columns of the matrix.
• The constants on the right of the equations form the last column, separated by a line.

Our system becomes the augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \2 & -3 & 2 & | & 4 \4 & 1 & -3 & | & 1\end{bmatrix}\]This matrix format makes it easier to apply row operations and move towards solving the system. Remember, the goal is to simplify the matrix and make the problem more manageable.

Triangular Form

Achieving a triangular form is a crucial step in solving systems of equations. It means transforming the matrix so that all elements below the main diagonal are zeros. This form is essential as it prepares the system for back substitution, our final solving step.
In our solution, the process was done by eliminating variables from each column, starting with the x-terms below the first row:

• Subtract 2 times the first row from the second row.
• Subtract 4 times the first row from the third row.

This yielded the intermediate matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & -3 & -7 & | & -7\end{bmatrix}\]Continuing, we eliminated the y-term in the third row by adding \(\frac{3}{5}\) of the second row to the third,
resulting in the triangular form:\[\begin{bmatrix}1 & 1 & 1 & | & 2 \0 & -5 & 0 & | & 0 \0 & 0 & -7 & | & -7\end{bmatrix}\]

Back Substitution

Once the matrix is in triangular form, back substitution is the final step. It's where we find the values of the variables, starting from the bottom row and working upwards.

• From the last equation: we have \(-7z = -7\) which leads to \(z = 1\).
• Next, we substitute \(z = 1\) into the second equation: \(-5y = 0\), giving \(y = 0\).
• Finally, substituting \(y = 0\) and \(z = 1\) into the first equation, we get \(x + 0 + 1 = 2\). Thus, \(x = 1\).

Back substitution allows for the neat solution of the system: \((x, y, z) = (1, 0, 1)\). It's like working backwards to reveal the answer, utilizing the neatly arranged zeros from the triangular form.