Problem 51 - Precalculus Mathematics for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{array}{r} x-y \quad+w=0 \\ 3 x \quad-z+2 w=0 \\ x-4 y+z+2 w=0 \end{array}\right.$$

Step-by-Step Solution

Answer

The general solution is \((x, y, z, w) = (s-t, s, 3s-t, t)\), where \(s, t \in \mathbb{R}\).

Step 1: Write the Equations in Matrix Form

The system of equations can be rewritten in matrix form as: \[\begin{pmatrix} 1 & -1 & 0 & 1 \ 3 & 0 & -1 & 2 \ 1 & -4 & 1 & 2 \end{pmatrix}\begin{pmatrix} x \ y \ z \ w \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]This sets up a system where we can apply row operations to simplify.

Step 2: Perform Gaussian Elimination

We start by eliminating the first element in the second and third row:- Subtract 3 times the first row from the second row to eliminate the \(x\)-term in the second row: \ \( R_2 = R_2 - 3R_1 \) \ \ \( (3, 0, -1, 2) - 3(1, -1, 0, 1) = (0, 3, -1, -1) \)- Subtract the first row from the third row to eliminate the \(x\)-term in the third row: \ \( R_3 = R_3 - R_1 \) \ \ \( (1, -4, 1, 2) - (1, -1, 0, 1) = (0, -3, 1, 1) \)

Step 3: Further Simplify with Row Operations

Now we have the system:\[\begin{pmatrix} 1 & -1 & 0 & 1 \ 0 & 3 & -1 & -1 \ 0 & -3 & 1 & 1 \end{pmatrix}\begin{pmatrix} x \ y \ z \ w \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]Add the second and third rows to eliminate the \(y\)-term in the third row: \ \( R_3 = R_3 + R_2 \) \ \ \( (0, -3, 1, 1) + (0, 3, -1, -1) = (0, 0, 0, 0) \)This suggests the system is dependent and has a free variable.

Step 4: Express Variables in Terms of Free Variables

The matrix simplifies to:\[\begin{pmatrix} 1 & -1 & 0 & 1 \ 0 & 3 & -1 & -1 \ 0 & 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \ y \ z \ w \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]The third row indicates dependence. Let's solve the system in terms of the free variable \(w\):- From the first row: \ \(x - y + w = 0 \rightarrow x = y - w\)- From the second row: \ \(3y - z - w = 0 \rightarrow z = 3y - w\)Now \(y\) and \(w\) are the free variables.

Step 5: Assign Parameters to Free Variables

Let's set: - \(y = s\) - \(w = t\)where \(s\) and \(t\) are parameters.Then substitute these into the expressions found earlier:- \(x = s - t\)- \(z = 3s - t\)This means the general solution for the system is:\[(x, y, z, w) = (s-t, s, 3s-t, t)\] for any real numbers \(s\) and \(t\).

Key Concepts

Gaussian elimination

Gaussian elimination is a systematic method used to solve systems of linear equations. It involves performing a series of row operations on a matrix until the matrix reaches its row-echelon form. The main objective is to simplify the system to make the solutions easier to find. Here’s how it typically works:

• Start by representing the system of equations as a matrix.
• Use row operations to zero out the coefficients below the pivot positions (the leading coefficients) in each column.
• Continue this process, moving from the top-left to the bottom-right of the matrix, until every column has a clear leading position.

In our exercise, we used Gaussian elimination by manipulating the matrix rows to ultimately find a simpler system. This process revealed dependencies between the variables, allowing us to identify free variables that could vary while maintaining a valid solution.

Matrix form

Writing a system of equations in matrix form is a powerful way to analyze and solve it. In matrix form, the system is expressed as an equation involving a coefficient matrix, a variable column vector, and a constant column vector:\[A \begin{pmatrix} x \ y \ z \ w \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]Here,

• The matrix \(A\) contains the coefficients of the variables from each equation.
• The vector \(x, y, z, w\) represents the variables we are solving for.
• The right side represents the constants from the equations, often zero in homogeneous systems.

Using matrix form streamlines the process of solving complex systems by allowing easy application of Gaussian elimination methods. This form displays the relationships between equations clearly, helping in identifying any potential solutions directly or through computational methods.

Free variables

Free variables occur in linear systems when there are fewer independent equations than unknowns. They lead to a system having infinitely many solutions because the solutions can "vary freely" within certain constraints set by the other variables. In the context of our exercise, once Gaussian elimination reduced the system, it revealed rows comprising entirely of zeros signifying dependence within the system.To handle free variables, we:

• Identify which variables can be freely set to a parameter (e.g., \(y = s\), \(w = t\) in solving the exercise).
• Express other variables in terms of these parameters, acknowledging their dependence on free variables.

Free variables are important because they illustrate the system's underlying structure and the existence of multiple solutions. In many practical applications, understanding the role of free variables can lead to insights about the degrees of freedom present in the system.