Problem 5 - Beginning and Intermediate Algebra

Question

Solve each system. \begin{aligned} x+3 y+z &=3 \\ 4 x-2 y+3 z &=7 \\ -2 x+y-z &=-1 \end{aligned}

Step-by-Step Solution

Answer

The solution to the given system of linear equations is \(x = 1, y = -1, z = 5\).

Step 1: Write the augmented matrix for the system

We will write the given system of linear equations as an augmented matrix: \(\begin{bmatrix} 1 & 3 & 1 &|& 3 \\ 4 & -2 & 3 &|& 7 \\ -2 & 1 & -1 &|& -1 \\ \end{bmatrix}\)

Step 2: Determine the row echelon form

To do this, we'll use Gaussian elimination to obtain zeroes in the lower-left corner of the matrix. We can start by swapping the first and second rows: \(\begin{bmatrix} 4 & -2 & 3 &|& 7 \\ 1 & 3 & 1 &|& 3 \\ -2 & 1 & -1 &|& -1 \\ \end{bmatrix}\) Next, perform the following row operations: \( Row \thinspace 2 \rightarrow Row \thinspace 2 - \frac{1}{4} Row \thinspace 1 \) \( Row \thinspace 3 \rightarrow Row \thinspace 3 + \frac{1}{2} Row \thinspace 1 \) After these operations, the matrix becomes: \(\begin{bmatrix} 4 & -2 & 3 &|& 7 \\ 0 & 3.5 & 0.25 &|& 1.25 \\ 0 & 0 & -0.5 &|& -2.5 \\ \end{bmatrix}\) Now, the matrix is in row echelon form.

Step 3: Determine the reduced row echelon form

Now, we will use back-substitution to get the matrix in reduced row echelon form (RREF). Perform the row operations: \( Row \thinspace 1 \rightarrow Row \thinspace 1 + \frac{2}{7} Row \thinspace 2 \) \( Row \thinspace 2 \rightarrow \frac{2}{7} Row \thinspace 2 \) \( Row \thinspace 3 \rightarrow -2 \thinspace Row \thinspace 3 \) After these operations, the matrix becomes: \(\begin{bmatrix} 4 & 0 & 3.5714 &|& 8.5714 \\ 0 & 2 & 0.5 &|& 1.5 \\ 0 & 0 & 1 &|& 5 \\ \end{bmatrix}\) Finally, perform the following operations: \( Row \thinspace 1 \rightarrow Row \thinspace 1 - 3.5714 \thinspace Row \thinspace 3 \) \( Row \thinspace 2 \rightarrow Row \thinspace 2 - 0.5 \thinspace Row \thinspace 3 \) Resulting in the matrix: \(\begin{bmatrix} 1 & 0 & 0 &|& 1 \\ 0 & 1 & 0 &|& -1 \\ 0 & 0 & 1 &|& 5 \\ \end{bmatrix}\)

Step 4: Interpret the results

Since the matrix is in reduced row echelon form, we can easily read off the values of x, y, and z from the matrix: x = 1 y = -1 z = 5 So the solution to the given system of linear equations is: \(x = 1, y = -1, z = 5\)

Key Concepts

System of Linear Equations

A system of linear equations involves multiple equations that share a common set of variables. In our example, the system given is three equations with three variables:

• \(x + 3y + z = 3\)
• \(4x - 2y + 3z = 7\)
• \(-2x + y - z = -1\)

The goal is to find a solution set of values for the variables \(x, y, z\) that satisfy all equations simultaneously. Such systems can describe various practical problems, such as intersecting lines or planes in physics or economics. Solving them requires methods that can handle multiple variables effectively.

Augmented Matrix

An augmented matrix is a crucial tool for solving systems of linear equations using matrix operations. It combines the coefficients of the system's equations with their constant terms into a single matrix. For our system, the augmented matrix looks like this:\[\begin{bmatrix}1 & 3 & 1 &|& 3 \4 & -2 & 3 &|& 7 \-2 & 1 & -1 &|& -1 \\end{bmatrix}\]In the augmented matrix, each row corresponds to one equation. The numbers before the vertical line (|) are the coefficients of the variables, and those after are the constants from each equation. This form is convenient for applying linear algebra techniques like Gaussian elimination.

Row Echelon Form

The row echelon form (REF) of a matrix simplifies the structure, making it easier to solve a system of equations. To achieve this form:

• All non-zero rows are above any rows of all zeroes.
• The leading coefficient (first non-zero number from the left) of a non-zero row is always to the right of the leading coefficient of the row above it.

Using Gaussian elimination, we transformed our initial augmented matrix into:\[\begin{bmatrix}4 & -2 & 3 &|& 7 \0 & 3.5 & 0.25 &|& 1.25 \0 & 0 & -0.5 &|& -2.5 \\end{bmatrix}\]This step, accomplished by using row operations, forms the basis for reaching the reduced row echelon form and ultimately solving the system.

Reduced Row Echelon Form

The reduced row echelon form (RREF) is a further simplification of a matrix, leading to a clear solution for the system of equations. Characteristics of RREF include:

• Each leading entry in a row is \(1\) (also known as a pivot), and is the only non-zero entry in its column.
• All zero rows, if any, are at the bottom.
• The leading entry of each non-zero row appears in a column to the right of the leading entry in the row above it.

For our system, performing additional transformations, the matrix was reduced to:\[\begin{bmatrix}1 & 0 & 0 &|& 1 \0 & 1 & 0 &|& -1 \0 & 0 & 1 &|& 5 \\end{bmatrix}\]From this form, the solution \(x = 1\), \(y = -1\), and \(z = 5\) can be easily read off, providing clear values for each variable that resolve the system.