Problem 21 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{array}{r} x \quad-4 z=1 \\ 2 x-y-6 z=4 \\ 2 x+3 y-2 z=8 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = 5\), \(y = 0\), \(z = 1\).

Step 1: Write the system of equations in matrix form

Write the given system of equations as a matrix equation, combining the coefficients and constants separately: \[\begin{bmatrix} 1 & 0 & -4 \ 2 & -1 & -6 \ 2 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 4 \ 8 \end{bmatrix}\]

Step 2: Perform row operations to get upper triangular form

Apply row operations to transform the coefficient matrix into an upper triangular form. This makes it easier to solve using back substitution:1. \(R_2 = R_2 - 2R_1\): \[\begin{bmatrix} 1 & 0 & -4 & | & 1 \ 0 & -1 & 2 & | & 2 \ 2 & 3 & -2 & | & 8 \end{bmatrix} \]2. \(R_3 = R_3 - 2R_1\): \[\begin{bmatrix} 1 & 0 & -4 & | & 1 \ 0 & -1 & 2 & | & 2 \ 0 & 3 & 6 & | & 6 \end{bmatrix} \]

Step 3: Eliminate further to achieve row-echelon form

Continue with row operations to further simplify the matrix:3. Multiply \(R_2\) by -1 to make the leading coefficient positive: \[\begin{bmatrix} 1 & 0 & -4 & | & 1 \ 0 & 1 & -2 & | & -2 \ 0 & 3 & 6 & | & 6 \end{bmatrix} \]4. Eliminate \(y\) from the third row using \(R_3 = R_3 - 3R_2\): \[\begin{bmatrix} 1 & 0 & -4 & | & 1 \ 0 & 1 & -2 & | & -2 \ 0 & 0 & 12 & | & 12 \end{bmatrix} \]

Step 4: Solve using back substitution

Now that the matrix is in upper triangular form, solve the equations starting from the last row upwards:1. From the last row: \(12z = 12\) implies \(z = 1\).2. Substitute \(z = 1\) into the second row: \(y - 2(1) = -2\) gives \(y = 0\).3. Substitute \(z = 1\) and \(y = 0\) into the first row: \(x - 4(1) = 1\) gives \(x = 5\).

Step 5: Verify and state the solution

Substitute \(x = 5\), \(y = 0\), \(z = 1\) back into the original equations to ensure they satisfy all three equations. The substitutions confirm all original equations are satisfied.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a shared set of unknowns. These equations function together to determine the values of these unknown variables. In our problem, the unknowns are \(x\), \(y\), and \(z\), and our task is to find values for them that satisfy all given equations simultaneously.

Typically, systems of equations can represent real-life scenarios, such as finding the intersection point of lines or planes in geometry. For linear equations like the ones you're working with, solutions might be a single point (for three variables), infinitely many solutions, or none if the lines or planes don't intersect in a meaningful way.

The primary methods for solving these systems include substitution, elimination, and matrix methods. Matrix methods are particularly powerful for complex systems, allowing for easy manipulation using computers or calculators.

Matrix Operations

Matrices provide a compact way to represent and manipulate systems of linear equations. They combine coefficients of the variables into a grid-like structure, allowing for easy arithmetic operations. In our scenario, matrix operations help simplify the equation system, making it easier to solve.

You start by constructing a coefficient matrix, a variable matrix, and a constant matrix from the system of equations. Each row in the matrix corresponds to one equation from the system. Once the system is in matrix form, you can perform row operations, which involve:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another row

These operations help to get the matrix into a more manageable form, like row-echelon or reduced row-echelon form, which is crucial for simplifying the process of finding solutions.

Using matrix methods makes the process of solving systems of equations efficient and clear, especially when dealing with larger systems.

Upper Triangular Matrix

In solving systems of equations through matrices, one key objective is to transform the coefficient matrix into an upper triangular matrix. An upper triangular matrix is a square matrix where all the entries below the main diagonal are zero.

Transforming a matrix into this form simplifies the solving process as it allows systematic back-substitution to derive variable values easily. This transformation involves strategic row operations, like those detailed in the steps of the solution. By methodically eliminating lower entries, you arrive at an upper triangular shape.

Once the matrix is upper triangular, you can start from the last equation (or row) and move upwards, substituting to find each variable's value step by step. This technique is often referred to as Gaussian elimination, a fundamental operation in linear algebra for solving systems of linear equations efficiently.