Problem 48 - Precalculus Mathematics for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{array}{rr} x+y-z-w= & 6 \\ 2 x+\quad z-3 w= & 8 \\ x-y+4 w= & -10 \\ 3 x+5 y-z-w= & 20 \end{array}\right.$

Step-by-Step Solution

Answer

One specific solution is \((x, y, z, w) = (9, 1, 2, 0)\).

Step 1: Write the System in Matrix Form

Express the system of equations in matrix form, \[ \begin{bmatrix} 1 & 1 & -1 & -1 \ 2 & 0 & 1 & -3 \ 1 & -1 & 0 & 4 \ 3 & 5 & -1 & -1 \end{bmatrix} \begin{bmatrix} x \ y \ z \ w \end{bmatrix} = \begin{bmatrix} 6 \ 8 \ -10 \ 20 \end{bmatrix} \] This is the system in the form of \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constants matrix.

Step 2: Use Gaussian Elimination to Simplify the System

Apply Gaussian elimination to the matrix \( A \). The first step is to make zeros below the first pivot by using row operations:1. Row 2 = Row 2 - 2·Row 1 2. Row 3 = Row 3 - Row 1 3. Row 4 = Row 4 - 3·Row 1This yields the matrix,\[ \begin{bmatrix} 1 & 1 & -1 & -1 \ 0 & -2 & 3 & -1 \ 0 & -2 & 1 & 5 \ 0 & 2 & -4 & 2 \end{bmatrix} \begin{bmatrix} x \ y \ z \ w \end{bmatrix} = \begin{bmatrix} 6 \ -4 \ -16 \ 2 \end{bmatrix} \]

Step 3: Further Simplify Using Gaussian Elimination

Continue with Gaussian elimination to make zeros below the second pivot in the second column:1. Row 4 = Row 4 + Row 2This results in the matrix, \[ \begin{bmatrix} 1 & 1 & -1 & -1 \ 0 & -2 & 3 & -1 \ 0 & -2 & 1 & 5 \ 0 & 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \ w \end{bmatrix} = \begin{bmatrix} 6 \ -4 \ -16 \ -2 \end{bmatrix} \]

Step 4: Simplify and Solve for the Variables

Using back substitution on the triangular matrix gives:- From Row 4: \( -z + w = -2 \) gives \( z = w + 2 \) - From Row 3: substitute \( z = w + 2 \) in \( -2y + z + 5w = -16 \) to get \( -2y + (w + 2) + 5w = -16 \).Solve to find \( y = \frac{w + 6}{2} \).- From Row 2: substitute \( z = w+2 \), plug into \( -2y + 3(w+2) - w = -4 \) to find \( y = \frac{w + 2}{2} \).- Row 1 gives \( x + y - (w+2) - w = 6 \), solve for x: \( x = 10 - y \).Thus, solving for each gives \( y = \frac{w+2}{2} \), \( z = w+2 \), and \( x = 10 - \frac{w+2}{2} \).

Step 5: Find a Specific Solution for Free Variable w

Choose a specific integer value for the free variable, \( w \). If \( w = 0 \), substituting back gives:- \( y = \frac{0+2}{2} = 1 \)- \( z = 0 + 2 = 2 \)- \( x = 10 - 1 = 9 \)Thus, the specific solution for \( w = 0 \) is \((x, y, z, w) = (9, 1, 2, 0)\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. The goal is to find values for these variables that make all the equations true at the same time. In mathematical terms, a system of linear equations can be written as: - \[ \begin{align*} \text{Equation 1:} & \quad a_1x + b_1y + c_1z + \ldots = d_1 \ \text{Equation 2:} & \quad a_2x + b_2y + c_2z + \ldots = d_2 \ \end{align*} \]- Equations such as the one given in the exercise form a system. Each equation can be graphically represented as a line or plane, and the intersection points of these lines or planes represent the solutions to the system. - Working with these types of problems involves:

• Understanding how each variable interacts with others through the given equations.
• Searching for a common solution set that satisfies all equations simultaneously.
• Using mathematical operations to transform and solve these systems efficiently.

Knowing how to deal with a system of equations is crucial in various fields like engineering, physics, and mathematics, as it lays the foundation for understanding complex real-world systems.

Matrix Form

Matrix form is a compact way to write systems of linear equations. When equations are converted into this form, it makes manipulation and solving using computational tools much easier. - To express a system in matrix form, the coefficients of the variables from each equation are written in a square matrix (known as the coefficient matrix) and the variables themselves in another matrix (the variable matrix). The constants from each equation are then arranged in yet another matrix, which we'll call the constant matrix. All together this looks like:- \[ A\mathbf{X} = \mathbf{B} \]- Where:

• \(A\) is the coefficient matrix.
• \(\mathbf{X}\) is the matrix of variables \((x, y, z, \ldots)\).
• \(\mathbf{B}\) is the constants matrix.

This transformation from equations to matrices allows for the use of efficient, systematic methods like Gaussian elimination to solve the system. Instead of handling multiple equations separately, the entire system is dealt with as a single entity, which reduces complexity and improves clarity.

Back Substitution

Back substitution is a method used to solve systems of linear equations once they have been converted to an upper triangular form using Gaussian elimination or another similar process.- In an upper triangular matrix, each non-zero row typically has more zeros to the left than the row directly above. This means that the bottom-most equation will have only one variable, which can be solved directly.

• Starting from the last equation, solve for the variable present in that equation.
• Substitute this found value back into the equation directly above it to solve the next variable.
• Continue this back substitution process upwards until all variables are found.

For example, if the bottom row gives \(-z + w = -2\), you can determine \(z\) in terms of \(w\). Then moving upwards, substitute \(z = w + 2\) into equations higher up, simplifying and solving for every variable step by step.- Back substitution is particularly powerful because it efficiently simplifies computation, making it ideal for these kinds of linear algebra problems. It also helps in verifying the integrity of solutions by gradually building from simpler equations to more complex ones.