Problem 34 - Precalculus: Mathematical for Calculus

Question

Use a calculator that can perform matrix operations to solve the system, as in Example 7. $$\left\\{\begin{array}{lr} x+\frac{1}{2} y-\frac{1}{3} z= & 4 \\ x-\frac{1}{4} y+\frac{1}{6} z= & 7 \\ x+y-z= & -6 \end{array}\right.$$

Step-by-Step Solution

Answer

Use matrix functions to find \(x = 6, y = -8, z = -10\).

Step 1: Express the System as a Matrix Equation

First, arrange the system of equations into matrix form. The system \(\begin{aligned} x + \frac{1}{2}y - \frac{1}{3}z &= 4, \ x - \frac{1}{4}y + \frac{1}{6}z &= 7, \ x + y - z &= -6 \end{aligned}\) can be written in the matrix equation form \(\mathbf{A}\mathbf{x} = \mathbf{b}\), where \(\mathbf{A} = \begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{3} \ 1 & -\frac{1}{4} & \frac{1}{6} \ 1 & 1 & -1 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}\), and \(\mathbf{b} = \begin{bmatrix} 4 \ 7 \ -6 \end{bmatrix}\).

Step 2: Use Matrix Inversion or Row Reduction

Enter the matrix \(\mathbf{A}\) and the vector \(\mathbf{b}\) into your calculator. You can use the calculator's functions to either find the inverse of \(\mathbf{A}\) (if it exists) and compute \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\), or perform row reduction to solve for \(\mathbf{x}\).

Step 3: Calculate the Solution

After applying the matrix operations, calculate the solution. For this system with matrix \(\mathbf{A}\) and \(\mathbf{b}\), solve to find the vector \(\mathbf{x}\). The calculator will give values for \(x\), \(y\), and \(z\).

Step 4: Verify the Solution

Substitute the solutions back into the original equations to verify correctness. This step ensures no calculation error occurred. Replace values of \(x\), \(y\), and \(z\) in each equation and check if both sides are equal.

Key Concepts

system of linear equations

A system of linear equations is essentially a collection of two or more linear equations that share the same set of variables. An equation like this often resembles \( y = mx + c \) in two dimensions, but extends to more variables in three or higher dimensions. In our exercise, the variables are \( x \), \( y \), and \( z \). Each line of the system represents a constraint. The main objective is to find values for these variables that will satisfy all equations in the system.

• For instance, the given system –
  \( x + \frac{1}{2}y - \frac{1}{3}z = 4 \),
  \( x - \frac{1}{4}y + \frac{1}{6}z = 7 \),
  \( x + y - z = -6 \) – depicts three planes intersecting in space, looking for a common point.

Systems like these are commonly encountered in Algebra and involve techniques such as substitution and elimination to solve for unknown variables. However, when we deal with larger systems or more complex equations, translating the system into a matrix form provides a more universal and efficient method for finding solutions.

• Understanding systems of linear equations is critical in various scientific disciplines due to their ability to model real-world phenomena, such as electrical circuits and economic forecasting.

matrix inversion

Matrix inversion is a pivotal concept when solving matrix equations, particularly useful when working with systems of linear equations. Let’s briefly consider what an inverse matrix is:

• Similar to how the inverse of a number \( n \) is \( 1/n \), the inverse of a matrix \( \mathbf{A} \), represented as \( \mathbf{A}^{-1} \), is a matrix which, when multiplied with \( \mathbf{A} \), yields the identity matrix \( \mathbf{I} \).

For solving a system of linear equations, matrix inversion allows us to find the solution vector \( \mathbf{x} \) by computing \( \mathbf{x} = \mathbf{A}^{-1}\mathbf{b} \). However, matrix inversion isn't applicable in every scenario. It only works if matrix \( \mathbf{A} \) is square (same number of rows and columns) and non-singular, meaning it has a determinant not equal to zero.

• When successfully applied, this method provides a straightforward path to finding solutions. However, in practice, inverting a matrix manually or computationally can be intensive.
• That said, calculators with matrix capabilities streamline this process significantly, making it a viable choice for many students tackling linear systems.

row reduction

Row reduction, also known as Gaussian elimination, is another powerful technique for finding solutions to systems of linear equations. This method involves performing operations on matrices to reach a simpler, often easier form to interpret results. The ultimate goal is to transform the matrix \( \mathbf{A} \) into its row echelon form or even further into reduced row echelon form.

• By reducing \( \mathbf{A} \) in this manner, you can directly arrive at the solution to the matrix equation without necessarily computing an inverse.
• Key operations include swapping rows, multiplying rows by a non-zero scalar, and adding or subtracting rows from each other to eliminate variables successively.

This method is effective and broadly applicable since it's not restricted by the requirement of a square or non-singular matrix like matrix inversion.

• Moreover, row reduction is the fundamental mechanism underpinning many computer algorithms for solving systems of equations, making it particularly relevant in computational settings.

By understanding and practicing this method, students gain a deeper insight into the structure and behavior of systems of linear equations.