Problem 25 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l} x+y+z=0 \\ 2 x-y+3 z=0 \\ x-z=0 \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Write the System as an Augmented Matrix

First, we need to express the given system of equations as an augmented matrix. The system of equations is written as: \( \begin{array}{c} x + y + z = 0 \ 2x - y + 3z = 0 \ x - z = 0 \end{array} \). This can be represented by the augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \2 & -1 & 3 & | & 0 \1 & 0 & -1 & | & 0 \\end{bmatrix}\]

Step 2: Apply Row Operations to Obtain Row Echelon Form

We need to create zeros below the first entry in the first column. Subtract row 1 from row 3 to make the entry below the first '1' zero.Then, subtract 2 times row 1 from row 2 to eliminate the first column's entry. The matrix becomes:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & -3 & 1 & | & 0 \0 & -1 & -2 & | & 0 \\end{bmatrix}\]

Step 3: Eliminate Second Column Entry

Next, we need to make the entry (2,2) equal to 1. Let's divide the second row by -3:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 1 & -\frac{1}{3} & | & 0 \0 & -1 & -2 & | & 0 \\end{bmatrix}\] Now, add row 2 to row 3 to make the entry below the leading 1 in the second column zero:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 1 & -\frac{1}{3} & | & 0 \0 & 0 & -\frac{7}{3} & | & 0 \\end{bmatrix}\]

Step 4: Solve for Third Variable Using Back Substitution

Since there is only one equation involving the variable \(z\) in the matrix, solve for \(z\) using the last row which gives: \(-\frac{7}{3}z = 0\) implying \(z = 0\).

Step 5: Substitute Back to Find Remaining Variables

Substitute \(z = 0\) back into the second row equation: \(y - \frac{1}{3}(0) = 0\), so \(y = 0\). Similarly, substitute \(y = 0\) and \(z = 0\) into the first equation: \(x + 0 + 0 = 0\), giving \(x = 0\).

Step 6: Collect Solution

The solution to the system of equations is \(x = 0\), \(y = 0\), and \(z = 0\). Therefore, the system has a unique solution.

Key Concepts

Augmented Matrix

In solving systems of linear equations, an **augmented matrix** plays a significant role. This is a way to express a linear system, simplifying the process and easing the transition to matrix operations. An augmented matrix comprises the coefficients of the variables and the constants from the equations, all in one compact form. For instance, given a system of equations:

• \( x + y + z = 0 \)
• \( 2x - y + 3z = 0 \)
• \( x - z = 0 \)

We write the system as:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \2 & -1 & 3 & | & 0 \1 & 0 & -1 & | & 0\end{bmatrix}\]Here, the part of the matrix before the vertical line represents the coefficients of the variables \(x, y,\) and \(z\), while the column after it is simply the constants from the equations. Augmented matrices provide a streamlined approach for systematic row operations in the solution process.

Row Operations

**Row operations** are fundamental in the manipulation of augmented matrices during Gaussian elimination. They enable transforming the matrix into a form that is easier to solve. There are three basic types of operations:

• **Swapping rows** – changing the order of the rows.
• **Multiplying a row** by a non-zero constant – changing all elements of a row by the same factor.
• **Adding or subtracting rows** – adding a multiple or the whole of one row to/from another row.

For example, in Gaussian elimination, we aim to form zeros below the pivot (leading diagonal) positions. Starting with our matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \2 & -1 & 3 & | & 0 \1 & 0 & -1 & | & 0\end{bmatrix}\]To eliminate the number below the first entry in the first column, we perform row operations:

• Subtract row 1 from row 3: yields zero in the desired spot.
• Subtract twice row 1 from row 2 to create another zero. This turns the matrix closer to a stepping stone for finding the solution.

Understanding these operations enables unlocking the mathematical method of reaching an increasingly simpler and more solvable matrix.

Back Substitution

Once an augmented matrix has been transformed into its **row echelon form**, the next step is **back substitution**. This is the process of solving for unknown values starting from the bottom of the matrix upward.

In the given solution, after the row operations, our matrix becomes:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 1 & -\frac{1}{3} & | & 0 \0 & 0 & -\frac{7}{3} & | & 0 \end{bmatrix}\]Here, the last row implies that: \(-\frac{7}{3}z = 0\). Solving this, we find \(z = 0\).
Next, substitute \(z = 0\) back into the second row equation, leading to \(y - \frac{1}{3}(0) = 0\), so \(y = 0\).
Finally, substituting both \(y = 0\) and \(z = 0\) into the first row equation gives \(x + 0 + 0 = 0\), resulting in \(x = 0\).

This orderly process ensures that each solution is systematically found, reducing possible errors, and confirming the system's full solution.