Problem 25 - College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Write the Augmented Matrix

First, we write the given system of equations as an augmented matrix. The system is: \[ \begin{align*} x+y+z &= 0 \ 2x-y+3z &= 0 \ x-z &= 0 \end{align*} \] This translates to the augmented matrix: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 2 & -1 & 3 & | & 0 \ 1 & 0 & -1 & | & 0 \end{bmatrix} \]

Step 2: Apply Row Operations

Start transforming the matrix to row-echelon form using Gaussian elimination. First, eliminate the first coefficient in the second and third rows: Subtract Row 1 from Row 3: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 2 & -1 & 3 & | & 0 \ 0 & -1 & -2 & | & 0 \end{bmatrix} \] Subtract 2 times Row 1 from Row 2: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & -3 & 1 & | & 0 \ 0 & -1 & -2 & | & 0 \end{bmatrix} \]

Step 3: Simplify the Second Row

Divide the second row by -3 to make the leading coefficient 1: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & -\frac{1}{3} & | & 0 \ 0 & -1 & -2 & | & 0 \end{bmatrix} \]

Step 4: Eliminate Coefficient Below Leaning 1 in Second Column

Add Row 2 to Row 3 to eliminate the coefficient below the leading 1 in the second column: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & -\frac{1}{3} & | & 0 \ 0 & 0 & -\frac{7}{3} & | & 0 \end{bmatrix} \]

Step 5: Back Substitution

The third row now implies that \(-\frac{7}{3}z = 0\) so \(z = 0\).Substituting \(z = 0\) in row 2 gives \(y = 0\).Substituting \(y = 0, z = 0\) into row 1 gives \(x = 0\).

Step 6: Solution Verification

Verify the solution by plugging \(x = 0\), \(y = 0\), and \(z = 0\) back into the original equations.\[ \begin{align*} x + y + z &= 0 \ 2x - y + 3z &= 0 \ x - z &= 0 \end{align*} \]All hold true with the solution \(x = 0\), \(y = 0\), \(z = 0\).

Key Concepts

Augmented Matrix

When faced with a system of linear equations such as:

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

An excellent way to manage these equations is through the use of an augmented matrix. This is a compact and organized way to express the system by aligning all coefficients and constants. The augmented matrix combines the coefficients of variables and the constants in one graphical representation.
For example, the matrix for the equations above would be:\[\begin{bmatrix} 1 & 1 & 1 & | & 0 \ 2 & -1 & 3 & | & 0 \ 1 & 0 & -1 & | & 0 \end{bmatrix}\]Here, each row corresponds to an equation, and each column corresponds to the coefficients of a variable or the constants on the right-hand side.
Using an augmented matrix helps streamline solving processes like Gaussian elimination by capturing all necessary data in a concise form.

Row Operations

Row operations are the key actions we apply to transform an augmented matrix during Gaussian elimination. These operations are elementary, yet powerful, allowing us to systematically simplify the matrix while preserving the solution set. There are three main types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting a multiple of one row from another

For instance, during our exercise, we performed operations like subtracting multiples of one row from another to eliminate specific coefficients.
This helps isolate variables column by column and aims at zeroing below pivotal numbers.For example, by subtracting Row 1 from Row 3, we eliminate the coefficient of \(x\) in Row 3, making the transformation:\[\begin{bmatrix} 1 & 1 & 1 & | & 0 \ 2 & -1 & 3 & | & 0 \ 0 & -1 & -2 & | & 0 \end{bmatrix}\]

Row-Echelon Form

Transforming a matrix into row-echelon form (REF) is a crucial step during Gaussian elimination. This form makes it easier to discern the values of variables through a structured triangular pattern of coefficients.
In row-echelon form, each non-zero row has a leading 1, and every leading 1 of a lower row appears to the right of the leading 1 of the row just above it. More importantly, all coefficients below these leading 1s are zero.
For instance, in our example, after applying necessary row operations, we achieved the matrix:\[\begin{bmatrix} 1 & 1 & 1 & | & 0 \ 0 & 1 & -\frac{1}{3} & | & 0 \ 0 & 0 & -\frac{7}{3} & | & 0 \end{bmatrix}\]This is row-echelon form. It simplifies back substitution, aiding in quickly finding the solution by addressing each row as an equation.

Back Substitution

Once a matrix has been transformed into row-echelon form, the final step in Gaussian elimination is back substitution. This process involves solving for variables starting from the bottom-most row upward.
In our case, beginning with \(-\frac{7}{3}z = 0\), we find that \(z = 0\). With \(z\) known, we substitute back into prior rows to find each preceding variable. For Row 2, this means solving for \(y\):\[0 \cdot y - \frac{1}{3} \cdot 0 = 0 \Rightarrow y = 0\]Continuing the process, substitute both \(y = 0\) and \(z = 0\) back into Row 1 to find \(x\):\[x + 0 + 0 = 0 \Rightarrow x = 0\]Back substitution is crucial for drawing exact values for variables once the system has been simplified. This step assures that all solutions satisfy the original set of equations.