Problem 91 - Precalculus with Limits

Question

In Exercises 91-94, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices. (a) \( \left\\{ \begin{array}{l} x - 2y + z = -6 \\ y - 5z = 16 \\ z = -3 \\ \end{array} \right. \) (b) \( \left\\{ \begin{array}{l} x + y - 2z = 6 \\ y + 3z = -8 \\ z = -3 \\ \end{array} \right. \)

Step-by-Step Solution

Answer

Yes, the two systems of linear equations yield the same solution, which is \( (-9, 1, -3) \).

Step 1: Solve System (a)

Substitute z=-3 into the first two equations, then system (a) becomes: \[\begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -5 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} -6 \\ 16 \\ -3 \end{bmatrix}.\] The solution is \((x, y, z) = (-9, 1, -3)\)

Step 2: Solve System (b)

Similarly, substitute z=-3 into the first two equations, then system (b) becomes: \[\begin{bmatrix} 1 & 1 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ -8 \\ -3 \end{bmatrix}.\] And the solution is \((x, y, z) = (-9, 1, -3)\)

Step 3: Comparison

Because the solutions for both systems are \((-9, 1, -3)\), the two systems of linear equations yield the same solution.

Key Concepts

Solving Systems Using Matrices

The process of solving systems of linear equations with matrices involves several steps, which transforms the system into a matrix equation. In the example provided, the use of matrices simplifies the process of solving the systems. By encoding the coefficients of the variables and the constants into matrix form, it becomes easier to manipulate the system and find the solutions.

For systems like (a) and (b), one key step is to use a technique called substitution when a variable can be isolated, as we see with the variable 'z' in both systems. Once 'z' is substituted, the remaining equations can be arranged into an augmented matrix that represents the system. This sets the stage for employing matrix operations to solve for 'x' and 'y'.

Matrix Representation

After substituting 'z', the equations form an augmented matrix which brings clarity to the problem. Matrices are incredibly powerful for system solving since they allow for row operations that can simplify the equations, eventually leading to an answer. These operations help in achieving an upper triangular form from where you can use back substitution to solve for the remaining variables. Such a structured approach ensures that we can crisply move from complex systems to straightforward solutions.

Matrix Algebra

Matrix algebra is a fundamental part of solving linear systems with matrices, and it's essential to understand the operations that can be performed on matrices. Key operations include matrix addition, multiplication, and the finding of inverses where applicable. However, for solving systems as in the exercise provided, we're particularly interested in row operations.

Row Operations

When solving linear systems, row operations are used to simplify matrices into a form that is easily solvable. These operations are: swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row. As these operations mirror what you can do with linear equations, they do not change the solution of the system.

The ultimate goal often is to reach Row-Echelon form or even Reduced Row-Echelon form, which organizes the matrix into a format where the solution to the system of equations becomes apparent. The consistency of matrix algebra rules makes it a reliable and systematic way to approach linear equations.

Substitution Method

The substitution method is a classical algebraic approach to solving systems of linear equations. It involves solving one of the equations for one variable in terms of the others and then substituting this expression into the remaining equation or equations.

Implementation

In our specific exercise, the substitution method is applied by first isolating 'z' in both system (a) and system (b), since 'z' is already given as -3. This value is then substituted back into the other equations, which simplifies the system to only two variables that need to be found, 'x' and 'y'.

Using substitution in tandem with matrices, like in the given step-by-step solution, represents a blend of algebraic and advanced numerical techniques. This exemplifies how different methods can be combined to achieve a more efficient solution process. While the substitution method is straightforward, its pairing with matrix operations can greatly accelerate finding solutions for larger and more complicated systems.