Problem 16 - Algebra and Trigonometry

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+y+z &=0 \\\\-x+2 y+5 z &=3 \\ 3 x-y &=6 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \( x = 1 \), \( y = -3 \), \( z = 2 \).

Step 1: Write the Augmented Matrix

First, convert the given system of linear equations into an augmented matrix. The system of equations is: \( x + y + z = 0 \), \( -x + 2y + 5z = 3 \), and \( 3x - y = 6 \). Thus, the augmented matrix is:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \-1 & 2 & 5 & | & 3 \3 & -1 & 0 & | & 6\end{bmatrix}\]

Step 2: Apply Row Operations to Achieve Row Echelon Form

Perform row operations to simplify the matrix. Add row 1 to row 2, and replace row 2 with this sum:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 3 & 6 & | & 3 \3 & -1 & 0 & | & 6\end{bmatrix}\]Next, replace row 3 with \(-3 \times\) row 1 plus row 3:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 3 & 6 & | & 3 \0 & -4 & -3 & | & 6\end{bmatrix}\]

Step 3: Further Simplify the Matrix to Row Echelon Form

Continue simplifying by performing row operations. Multiply row 2 by \( \frac{1}{3} \) to make the leading coefficient 1:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 1 & 2 & | & 1 \0 & -4 & -3 & | & 6\end{bmatrix}\]Next, eliminate the \(-4\) in row 3 using row 2: Add \(4 \times\) row 2 to row 3:\[\begin{bmatrix}1 & 1 & 1 & | & 0 \0 & 1 & 2 & | & 1 \0 & 0 & 5 & | & 10\end{bmatrix}\]

Step 4: Solve for Variables Using Back Substitution

Now that the matrix is in row-echelon form, solve for variables by back substitution. From the last row: \[ 5z = 10 \Rightarrow z = 2 \]Substitute \( z = 2 \) into the second row equation:\[ y + 2(2) = 1 \Rightarrow y + 4 = 1 \Rightarrow y = -3 \]Substitute \( y = -3 \) and \( z = 2 \) into the first row equation:\[ x - 3 + 2 = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1 \]

Step 5: Verify the Solution

Substitute \( x = 1 \), \( y = -3 \), and \( z = 2 \) back into the original equations to verify:1. \( 1 - 3 + 2 = 0 \Rightarrow 0 = 0 \)2. \( -1 + 2(-3) + 5(2) = 3 \Rightarrow -1 - 6 + 10 = 3 \Rightarrow 3 = 3 \)3. \( 3(1) - (-3) = 6 \Rightarrow 3 + 3 = 6 \Rightarrow 6 = 6 \)The solution satisfies all equations.

Key Concepts

Augmented Matrix

In solving linear equations, we often use matrices to simplify the process. An augmented matrix is a useful tool that represents a system of linear equations. The matrix consists of the coefficients of the variables and the constants from the equations in a compact form. For the given system, the augmented matrix is built from the coefficients and constants:

• First row: the coefficients of the first equation, followed by its constant: \(1, 1, 1 | 0\).
• Second row: the coefficients of the second equation: \(-1, 2, 5 | 3\).
• Third row: the coefficients of the third equation: \(3, -1, 0 | 6\).

This representation allows us to handle the equations systematically and perform manipulations effectively.

Row Operations

Once we have an augmented matrix, we can perform row operations to simplify it. These operations are essential to solve linear equations and include:

• Swapping two rows.
• Multiplying a row by a non-zero constant.
• Adding or subtracting a multiple of one row to another.

These operations help us manipulate the matrix step-by-step toward a simpler form, known as row echelon form. For the given problem, we smooth the solution by adjusting the augmented matrix. By making calculated row additions and multiplications, we progressively simplify the system until reaching a form that lends easily to solution finding.

Row Echelon Form

Row echelon form is an intermediate step in solving systems of equations using matrices. This simplified form has certain characteristics that make it easier to solve using back substitution:

• All non-zero rows are above any rows of all zeros.
• The leading coefficient (first non-zero number from the left, also known as the pivot position) of a non-zero row is always to the right of the leading coefficient of the row just above it.
• All entries in a column below a leading coefficient are zeros.

For our system, we've achieved row echelon form, which allows us to move on to the final step of back substitution to find the solutions for each variable.

Back Substitution

Back substitution is used after reducing a system to row echelon form, allowing us to solve for the unknown variables starting from the last non-zero row of the matrix. The process involves:

• Solving the last equation in the system for its variable.
• Substituting that solution back into the equations above it.
• Continuing this process row by row until you reach the top, determining the value of each variable sequentially.

In our example, we started with solving for \(z\) from the last row, then used that value to find \(y\) from the second row, and finally solved for \(x\) using the first row. This method allows step-by-step, straightforward solving of each variable in turn.