Problem 32 - College Algebra with Modeling and Visualization

Question

Use Gaussian elimination with backward substitution to solve the system of linear equations. Write the solution as an ordered pair or an ordered triple whenever possible. $$ \begin{aligned} x+3 y-2 z &=-4 \\ 2 x+6 y+z &=-3 \\ x+y-4 z &=-2 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \((4, -2, 1)\).

Step 1: Convert System to Augmented Matrix

First, let's write the system of equations as an augmented matrix:\[\begin{bmatrix}1 & 3 & -2 & | & -4 \2 & 6 & 1 & | & -3 \1 & 1 & -4 & | & -2 \\end{bmatrix}\]

Step 2: Perform Row Operations to Get Upper Triangular Matrix

We want to get zeros below the leading 1 in the first column. We can subtract the first row from the third row and subtract twice the first row from the second row to get:\[\begin{bmatrix}1 & 3 & -2 & | & -4 \0 & 0 & 5 & | & 5 \0 & -2 & -2 & | & 2 \\end{bmatrix}\]

Step 3: Simplify Further to Upper Triangular Form

Next, let's simplify the third row by dividing by -2 and simplifying further:\[\begin{bmatrix}1 & 3 & -2 & | & -4 \0 & 0 & 5 & | & 5 \0 & 1 & 1 & | & -1 \\end{bmatrix}\]

Step 4: Back Substitute to Find Solution

Starting from the last row, we use back substitution. The third row gives us \(y + z = -1\). The second row simplifies to \(5z = 5\), giving \(z = 1\). Substitute \(z = 1\) into \(y + z = -1\) to solve for \(y\), obtaining \(y = -2\). Substitute \(y = -2\) and \(z = 1\) into the first equation to find \(x\): \(x + 3(-2) - 2(1) = -4\)which simplifies to \(x = 4\).

Step 5: Final Solution

The solution to the system, in ordered triple form, is \((x, y, z) = (4, -2, 1)\).

Key Concepts

Linear Equations

Linear equations are mathematical expressions involving constants and variables raised to the first power. Each term is a monomial. These equations describe straight lines when graphed, hence the name 'linear'. Linear equations can consist of multiple variables, forming a system of equations, such as in the exercise above.

In this exercise, our system has three linear equations:

• Equation 1: \( x + 3y - 2z = -4 \)
• Equation 2: \( 2x + 6y + z = -3 \)
• Equation 3: \( x + y - 4z = -2 \)

Solving a system of linear equations involves finding the values of the variables that make all the equations true at the same time.

We use techniques like Gaussian Elimination to find the solution, as these equations are interconnected, and a change in one variable affects the others. Gaussian elimination converts the system into a form that makes it easier to solve, by performing operations that simplify the original equations.

Back Substitution

Back substitution is a technique used to find the solutions of a system of equations after it has been transformed into an upper triangular matrix through techniques like Gaussian elimination.

This process involves:

• Starting from the last equation, which should involve only one variable once the system is in triangular form.
• Solving for that variable first, as it's directly given by the last row in the matrix.
• Using the solved value to find other variables by plugging it back into preceding equations.

In the example provided, the last row simplifies to \( y + z = -1 \), allowing us to find \( y \) and \( z \). The second-to-last equation helps us find \( z \), and substituting back to the upper rows enables finding \( x \). This tops-down approach is efficient and ensures all variables are found systematically.

Augmented Matrix

An augmented matrix is a key concept in solving systems of linear equations. It is a matrix that includes the coefficients of the variables as well as the constants from the right sides of the equations. This arrangement helps in systematically applying row operations to simplify and solve the system.

The augmented matrix for the exercise's system of equations looks like this:

• First row: \( [1, 3, -2 | -4] \)
• Second row: \( [2, 6, 1 | -3] \)
• Third row: \( [1, 1, -4 | -2] \)

The vertical bar separates the coefficients of the variables from the constants, aiding visualization of the system. This matrix allows us to use operations like row swapping and scaling to transform the system into an easier form, ultimately solving for the variables using methods like Gaussian elimination followed by back substitution.