Problem 105 - Precalculus

Question

Solve the system of equations using an augmented matrix. $$ \begin{aligned} 3 x-2 y+z &=-1 \\ x+y-z &=3 \\ 2 x-y+3 z &=0 \end{aligned} $$ Solution: \(\begin{aligned} \text { Step } 1: \text { Write the system as an } &\left[\begin{array}{rrr|r}3 & -2 & 1 & -1 \\ 1 & 1 & -1 & 3 \\ 2 & -1 & 3 & 0\end{array}\right] \\ \text { augmented matrix. } & 1 \end{aligned}\) \(\begin{aligned} \text { Step 2: Reduce the matrix using } &\left[\begin{array}{lll|l}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0\end{array}\right] \\ \text { Gaussian elimination. } & 0 \end{aligned}\) Step 3: Identify the answer: Row 3 is inconsistent \(1=0\) therefore there is no solution. This is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 2 \), \( z = 0 \). The system is consistent and solvable.

Step 1: Write the System as an Augmented Matrix

Convert each equation from the system into rows of an augmented matrix. For the system given, we construct the following augmented matrix: \[\begin{bmatrix} 3 & -2 & 1 & \vert & -1 \ 1 & 1 & -1 & \vert & 3 \ 2 & -1 & 3 & \vert & 0 \end{bmatrix}\]

Step 2: Apply Gaussian Elimination

Perform row operations to transform the matrix into row-echelon form:1. Make the first entry of the first row 1 by dividing the entire first row by 3.\[\begin{bmatrix} 1 & -\frac{2}{3} & \frac{1}{3} & \vert & -\frac{1}{3} \ 1 & 1 & -1 & \vert & 3 \ 2 & -1 & 3 & \vert & 0 \end{bmatrix}\]2. Eliminate x terms from rows 2 and 3 by subtracting appropriate multiples of row 1.

Step 3: Continue Gaussian Elimination

Complete the row-elimination process:1. Subtract the first row from the second row to eliminate x from the second row.2. Subtract twice the first row from the third row to eliminate x from the third row.The intermediate matrix should look like this:\[\begin{bmatrix} 1 & -\frac{2}{3} & \frac{1}{3} & \vert & -\frac{1}{3} \ 0 & \frac{5}{3} & -\frac{4}{3} & \vert & \frac{10}{3} \ 0 & \frac{1}{3} & \frac{7}{3} & \vert & \frac{2}{3} \end{bmatrix}\]

Step 4: Simplify to Row-Echelon Form

Continue simplifying the matrix to make it easier to solve:1. Multiply row 2 by \( \frac{3}{5} \) to make the leading coefficient of row 2 equal to 1.2. Eliminate y-term from row 3 using row 2.The matrix transforms into:\[\begin{bmatrix} 1 & -\frac{2}{3} & \frac{1}{3} & \vert & -\frac{1}{3} \ 0 & 1 & -\frac{4}{5} & \vert & 2 \ 0 & 0 & 3 & \vert & 0 \end{bmatrix}\]

Step 5: Back Substitution to Find Solutions

By inspection, solve from the bottom up:1. From row 3: \( 3z = 0 \) implies \( z = 0 \).2. Substitute \( z = 0 \) into row 2: \( y - \frac{4}{5}(0) = 2 \) implies \( y = 2 \).3. Substitute \( y = 2 \) and \( z = 0 \) into row 1: \( x - \frac{2}{3}(2) + \frac{1}{3}(0) = -\frac{1}{3} \), solving gives \( x = 1 \).

Step 6: Confirm Solution Consistency

Since the elimination leads to valid expressions without contradictions, and each variable has a single unique solution, the system is consistent.The solution for the system is: \( x = 1 \), \( y = 2 \), \( z = 0 \).

Key Concepts

Gaussian Elimination

Gaussian elimination is a powerful method to solve systems of linear equations. It involves transforming a system of equations into simpler forms using a series of operations, making it easier to find the values of the variables. This process primarily uses three types of row operations:

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

By systematically applying these operations, you can simplify the system into either row-echelon form or reduced row-echelon form. These forms help visually illustrate the solution of the system or indicate if no solution exists.

In our example, Gaussian elimination helped us transform the initial augmented matrix into a simpler form, from which we could extract the solutions for the variables. It's important to carry out each step carefully, as any small mistake can affect the final solution.

System of Equations

A system of equations comprises multiple equations that share the same set of variables. Solving such systems involves finding a set of values for the variables that satisfy all equations simultaneously.

Systems of equations can have one solution, infinitely many solutions, or no solution at all:

• A unique solution indicates that there is just one set of values for the variables that satisfies all equations.
• Infinitely many solutions suggest that there is more than one way to satisfy the equations, typically when the equations are dependent.
• No solution means that there is no possible way to satisfy all equations simultaneously, often due to the equations being contradictory.

In the exercise provided, we initially assumed there was no solution due to a misstep, but upon revisiting the steps, we found a unique solution. This highlights the importance of following the correct procedures when working with augmented matrices.

Row-Echelon Form

Row-echelon form (REF) is a type of matrix representation that simplifies solving systems of equations. In REF, the matrix has the following characteristics:

• All non-zero rows are above any rows of all zeros.
• The leading coefficient (first non-zero number from the left, also called a pivot) of a non-zero row is always to the right of the leading coefficient of the row above it.
• The leading coefficient in any non-zero row is 1.

This representation helps simplify back substitution, where you solve for the variables starting from the last equation up to the first.

In our exercise, transforming the system into REF allowed us to solve for the variables efficiently, employing back substitution to ascertain the values of the variables. It's a crucial step in the process, and achieving it correctly is key to ensuring the accuracy of the solutions.