Problem 81 - College Algebra and Calculus: An Applied Approach

Question

Determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices. (a) \(\left\\{\begin{aligned} x-4 y+5 z &=27 \\ y-7 z &=-54 \\ z &=8 \end{aligned}\right.\) (b) \(\left\\{\begin{aligned} x-6 y+z &=15 \\ y+5 z &=42 \\ z &=8 \end{aligned}\right.\)

Step-by-Step Solution

Answer

Yes, the two systems of equations have the same solutions, which are \(x = 113\), \(y = -16\), and \(z = 8\).

Step 1: Convert to Matrix Form

Convert the two systems of equations into matrices. The terms on the left-hand side of the equal sign represent the coefficients in the matrix. The terms on the right-hand side of the equal sign, which are solution of each equation, will be put into a separate matrix. So, the first system of equations becomes matrix A: \[ \begin{array}{ccc|c} 1 & -4 & 5 & 27 \ 0 & 1 & -7 & -54\ 0 & 0 & 1 & 8\end{array} \], and the second one becomes matrix B: \[ \begin{array}{ccc|c} 1 & -6 & 1 & 15 \ 0 & 1 & 5 & 42 \ 0 & 0 & 1 & 8\end{array} \].

Step 2: Solve the Matrices

Now solve the matrices independently to find the values for variables x, y, and z. For matrix A, since it's already in Row-Echelon form, start from the bottom to solve for z, then substitute z into the second equation to solve for y, and finally substitute y and z into the first equation to get x. Thus, \(x = 113\), \(y = -16\), and \(z = 8\). For matrix B, proceed with the same method for solving the matrix. Here too, (\(x = 113\), \(y = -16\), \(z = 8\)).

Step 3: Compare Solutions

The solution for both system of equations are identical. Thus, the two systems yield the same solution.

Step 4: Conclusion

The analysis of both system of linear equations confirms that they yield the same solution, which is \(x = 113\), \(y = -16\), and \(z = 8\).

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. In the context of linear equations, these systems are composed of equations that graph as straight lines. Key to solving these systems is finding the values of the variables that satisfy all the equations simultaneously.

The examples given in the exercise involve two systems of equations. Each system has three equations and three unknowns: \(x\), \(y\), and \(z\). For each system, the goal is to find one unique solution that fits all equations.
When solving these systems by hand or using matrices, consistently working through each equation ensures the solution is correct and aligns across the entire system.

Matrix Form

Matrix form is a mathematical representation where the system of linear equations is expressed using a matrix. Matrices help in simplifying and organizing the data of equations, allowing them to be manipulated more easily.

To convert a system of equations into matrix form, the coefficients of each variable are placed into a matrix. For instance, the system's coefficients create a 3x3 matrix, and the constants on the right side of the equal sign form a separate column matrix.

• For example, the system \(x-4y+5z=27\), \(y-7z=-54\), \(z=8\), becomes a larger augmented matrix:
  1. \(1, -4, 5 | 27\)
  2. \(0, 1, -7 | -54\)
  3. \(0, 0, 1 | 8\)

This format enables the use of techniques like row operations to solve the systems efficiently.

Row-Echelon Form

Row-echelon form is a specific way of arranging a matrix where each leading coefficient (the first non-zero number from the left) of a row is at least one column to the right of the leading coefficient of the row above it. Additionally, all non-zero rows are above any rows of all zeros.

Achieving row-echelon form simplifies solving linear equations using back substitution. Starting from the bottom row (which typically represents a simpler equation), one can solve the variables and work upwards.

• For example, in matrix A, due to the arrangement:
  1. The bottom row shows \(z = 8\) directly.
  2. Using \(z = 8\), solve the second row for \(y\).
  3. Substitute \(y\) and \(z\) into the first equation to find \(x\).

This structured method can effectively clarify and solve systems of linear equations.

Solution of Linear Systems

The solution of a linear system is the set of values that satisfy all equations simultaneously. In practical terms, these are the values you substitute back into the original equations to check if they hold true.

The solution can either be unique, infinite, or nonexistent.

• For a unique solution, one set of values satisfies all the original equations, as seen in the exercise where \((x = 113, y = -16, z = 8)\).
• It is important to verify the solution by substituting the values back into the original equations.

When both systems of equations in the exercise yield the same values for \(x\), \(y\), and \(z\), this confirms that the solutions are consistent and valid for both systems.