Problem 12 - Precalculus: Functions and Graphs

Question

Use matrices to solve the system. $$\left\\{\begin{array}{r} x+y-2 z=0 \\ x-y-4 z=0 \\ y+z=0 \end{array}\right.$$

Step-by-Step Solution

Answer

The system's solution is \((x, y, z) = (3t, -t, t)\), parameterized by \(t\).

Step 1: Write the System as a Matrix Equation

First, let's express the given system of equations in matrix form. The system can be written as:\[ \begin{align*}1x + 1y - 2z &= 0 \1x - 1y - 4z &= 0 \0x + 1y + 1z &= 0 \\end{align*} \]This can be expressed in a matrix form as: \( A \cdot \mathbf{X} = \mathbf{B} \) where:\[A = \begin{bmatrix} 1 & 1 & -2 \ 1 & -1 & -4 \ 0 & 1 & 1 \end{bmatrix}, \\mathbf{X} = \begin{bmatrix} x \y \z \end{bmatrix}, \and \ \mathbf{B} = \begin{bmatrix} 0 \0 \0 \end{bmatrix}.\]

Step 2: Form the Augmented Matrix

Construct the augmented matrix by combining matrix \( A \) with vector \( \mathbf{B} \). The augmented matrix is:\[\left[\begin{array}{ccc|c}1 & 1 & -2 & 0 \1 & -1 & -4 & 0 \0 & 1 & 1 & 0\end{array}\right]\]

Step 3: Perform Row Operations to Reach Row-Echelon Form

Apply row operations to convert the augmented matrix to row-echelon form:1. Subtract the first row from the second row:\[\left[\begin{array}{ccc|c}1 & 1 & -2 & 0 \0 & -2 & -2 & 0 \0 & 1 & 1 & 0\end{array}\right]\]2. Add the second row to the third row:\[\left[\begin{array}{ccc|c}1 & 1 & -2 & 0 \0 & -2 & -2 & 0 \0 & -1 & -1 & 0\end{array}\right]\]

Step 4: Simplify the Matrix Further

Continue simplifying to isolate variables:1. Multiply the second row by -1/2:\[\left[\begin{array}{ccc|c}1 & 1 & -2 & 0 \0 & 1 & 1 & 0 \0 & -1 & -1 & 0\end{array}\right]\]2. Add the second row to the third row:\[\left[\begin{array}{ccc|c}1 & 1 & -2 & 0 \0 & 1 & 1 & 0 \0 & 0 & 0 & 0\end{array}\right]\]

Step 5: Interpret the Resulting Matrix

The final row-echelon form matrix is:\[\left[\begin{array}{ccc|c}1 & 1 & -2 & 0 \0 & 1 & 1 & 0 \0 & 0 & 0 & 0\end{array}\right]\]This represents the system of equations:\[ x + y - 2z = 0 \y + z = 0 \\]

Step 6: Solve the System of Equations

From the second equation, solve for \( y \):\[ y = -z \]Substitute \( y = -z \) in the first equation:\[ x + (-z) - 2z = 0 \ x - 3z = 0 \ x = 3z \]Let \( z = t \) (a parameter), then \( x = 3t \) and \( y = -t \).

Step 7: Write the Solution as a Parameter Set

The solution can be expressed as:\[ \begin{bmatrix} x \y \z \end{bmatrix} = \begin{bmatrix} 3t \-t \t \end{bmatrix} \] for any real number \( t \).

Key Concepts

Matrix Equations

Understanding matrix equations is essential for solving systems of linear equations using matrices. A matrix equation is a compact way of representing a system of equations. Instead of dealing with separate equations, we combine them into a matrix form. This involves a matrix of coefficients, a vector of variables, and a vector representing the constants.

• The matrix of coefficients, often denoted by \( A \), includes all the coefficients of the variables from the equations.
• The variables form a column vector, \( \mathbf{X} \), typically representing \( x, y, z, \) etc.
• The constants form another column vector, \( \mathbf{B} \), which contains the right-hand side of each equation in the system.

When we have a system like this: \[ \begin{align*} 1x + 1y - 2z &= 0 \ 1x - 1y - 4z &= 0 \ 0x + 1y + 1z &= 0 \end{align*} \]It can be rewritten as \( A \cdot \mathbf{X} = \mathbf{B} \), providing a structured and organized way to solve these equations using matrix techniques, such as row operations.

Row Operations

In the journey of solving matrix equations, row operations play a pivotal role. They're the mathematical maneuvers we use to simplify matrices and reveal the solutions to our systems of equations. There are three fundamental types of row operations:

• Swapping Rows: We can swap two rows without changing the solutions of the matrix. This is useful when you want to rearrange the equations logically.
• Multiplying a Row by a Nonzero Scalar: Any row can be multiplied by a non-zero number. This helps in normalizing the matrix rows to achieve a clearer form.
• Addition/Subtraction of Rows: You can add or subtract rows from each other. This is most commonly used to eliminate variables and simplify the matrix.

For example, with the system at hand, inter-row operations such as subtracting or adding rows helped in simplifying the matrix further and eventually aims to reach it to its simplest form, called the row-echelon form.

Row-Echelon Form

Reaching the row-echelon form of a matrix is a strategic goal in solving systems of equations through matrices. In this form:

• All non-zero rows are above rows of all zeros.
• The leading coefficient (or pivot) of a non-zero row is always strictly to the right of the leading coefficient of the row above it.
• The leading entry in each non-zero row is 1.

To track our progress, it's useful to perform row operations until the matrix aligns with these conditions. Once you've reached the row-echelon form, the matrix gives a straightforward view of the solution as the first step toward solving the system. It helps in simplifying equations making them easier to solve. For instance, in our exercise, converting to the row-echelon form enabled us to write the system as simplified equations and easily solve it for variable values, unveiling any parameter solutions present.