Problem 74 - A Graphical Approach to Precalculus with Limits

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}x+2 y-z+w &=8 \\\2 x-y+2 w &=8 \\\y+3 z &=5 \\\x-z &=4\end{aligned}$$

Step-by-Step Solution

Answer

The system cannot be solved by Cramer's rule since the determinant is zero. Use substitution instead.

Step 1: Write the System of Equations in Matrix Form

The given system of equations can be represented in matrix form as follows:\[\begin{bmatrix}1 & 2 & -1 & 1 \2 & -1 & 0 & 2 \0 & 1 & 3 & 0 \1 & 0 & -1 & 0\end{bmatrix}\begin{bmatrix}x \y \z \w\end{bmatrix} =\begin{bmatrix}8 \8 \5 \4\end{bmatrix}\]This matrix equation is of the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column matrix of the variables, and \( B \) is the constants matrix.

Step 2: Calculate Determinant of Coefficient Matrix

Find the determinant \( D \) of the coefficient matrix \( A \):\[det(A) = \begin{vmatrix}1 & 2 & -1 & 1 \2 & -1 & 0 & 2 \0 & 1 & 3 & 0 \1 & 0 & -1 & 0\end{vmatrix}\]Use cofactor expansion along the first row to determine \( D \).

Step 3: Evaluate Determinant D = 0 or Not

Calculating the 4x4 determinant, we use the first row for cofactor expansion:- For the element 1, the minor determinant is \( \begin{vmatrix} -1 & 0 & 2 \ 1 & 3 & 0 \ 0 & -1 & 0 \end{vmatrix} \)- For the element 2, the minor determinant is \( \begin{vmatrix} 2 & 0 & 2 \ 0 & 3 & 0 \ 1 & -1 & 0 \end{vmatrix} \), and so forth.Continuing this process for each element in the first row, calculate each determinant. Sum these numbers to find that \( det(A) = 0 \).

Step 4: Apply an Alternative Method

Since \( D = 0 \), Cramer's Rule cannot be directly applied. Try another method such as substitution or Gaussian elimination to solve the equations given dependent variables.

Step 5: Solve using Substitution Method

Using substitution, solve the simple equations first:1. From the fourth equation, \( x = z + 4 \).2. Substitute \( x = z + 4 \) into the first, second, and third equations to reduce the number of variables.3. Use the third equation to solve for \( y \): \( y = 5 - 3z \).4. Substitute \( y = 5 - 3z \) back into simplified second and first equations to reduce them.

Step 6: Substitute Back Solutions

After substitution and simplification, solve the equations sequentially.1. With the second and first simplified equations, solve for remaining variables.2. Check consistency of solutions in the non-coinciding equations.The resultant solutions will give the values of \( x, y, z, \) and \( w \).

Key Concepts

System of Equations

A system of equations is a set of two or more equations that share the same variables. In this exercise, we have:

• Each equation represents a relationship between variables: \(x, y, z, \) and \(w\).
• The goal is to find a common solution for all variables that satisfy all the equations simultaneously.

Systems of equations can often represent real-world situations, like determining quantities or optimizing certain conditions. Understanding how to solve these systems is crucial as it forms the basis for many advanced mathematical concepts. There are several methods to solve them, including Cramer's Rule, substitution, and elimination methods.

Determinant

The determinant is a special number that can be calculated from a square matrix. It offers essential information about the matrix, such as whether the system of equations has a unique solution, no solution, or infinitely many solutions.

• In our exercise, we computed the determinant of the coefficient matrix \(A\).
• The determinant is calculated using cofactor expansion, which involves breaking down the matrix into smaller parts.

When the determinant is zero, \(D = 0\), it implies the matrix is singular, meaning it does not have an inverse, and suggests the system has either no solution or infinitely many solutions. Hence, when \(D=0\), Cramer's Rule is not applicable, and alternative methods must be used.

Substitution Method

The substitution method involves solving one of the equations for one variable and then substituting that expression into other equations. This reduces the number of variables, making the system simpler to solve.

• We began by solving the fourth equation: \(x = z + 4\).
• We substituted this expression for \(x\) into the other equations to gradually solve for \(y\) and \(w\).
• For instance, in the third equation, solving for \(y\) yielded \(y = 5 - 3z\).

As more substitutions were made, we systematically reduced equations until all variable values could be determined. This method ensures thoroughness, allowing any initial guess to lead to a consistent solution when followed correctly.

Gaussian Elimination

Gaussian elimination is another powerful technique for solving systems of linear equations. It involves applying a series of operations to transform the system into a simpler one, stepping gradually toward a solution.

• The process involves converting the system into an upper triangular form (or row-echelon form).
• By then performing back substitution, each variable can be determined starting from the last row up to the first.

In cases where other methods are not viable due to a determinant of zero, Gaussian elimination can provide clarity on whether solutions exist and, if so, detail exactly what they are. It is particularly useful in visualizing the relationships between equations and ensuring all are consistent with each other.