Problem 64 - Precalculus with Limits

Question

In Exercises 57-64, (a) write the system of linear equations as a matrix equation, \(AX\ =\ B\), and (b) use Gauss-Jordan elimination on the augmented matrix \([A\ \vdots\ B]\) to solve for the matrix \(X\). \(\begin{cases} x_1 - x_2 + 4x_3= 17 \\ x_1 + 3x_2 = -11 \\ -6x_2 + 5x_3 = 40 \end{cases}\)

Step-by-Step Solution

Answer

The solutions to the system of equations are \(x_1 = 1/3\), \(x_2 = 2/3\) and \(x_3 = 4\).

Step 1: Writing as a matrix equation

First, we'll express the given system of equations in the matrix form \(AX = B\). The matrix \(A\) will contain the coefficients of the variables, \(X\) is the column matrix of variables and \(B\) is the column matrix of constants. Hence, we have \[A = \begin{bmatrix} 1 & -1 & 4 \ 1 & 3 & 0 \ 0 & -6 & 5 \end{bmatrix}\], \[X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} \] and \[B = \begin{bmatrix} 17 \ -11 \ 40 \end{bmatrix}\]. Thus, the system is represented as \[AX = B\]

Step 2: Set up the augmented matrix

Form the augmented matrix \([A | B]\) as follows: \[\begin{bmatrix} 1 & -1 & 4 & 17 \ 1 & 3 & 0 & -11 \ 0 & -6 & 5 & 40 \end{bmatrix}\]

Step 3: Gauss-Jordan elimination

Begin by subtracting the first row from the second to get a zero in the first column of the second row. After this, multiply the third row by -1/6 to get a positive coefficient for \(x_2\). Lastly, add 3 times of the new third row to the second row to get a zero below the one in the second column. The augmented matrix \([A | B]\) after these steps becomes \[\begin{bmatrix} 1 & -1 & 4 & 17 \ 0 & 4 & -4 & -28 \ 0 & 1 & -5/6 & -20/3 \end{bmatrix}\] Now, add the second row to the first row and subtract 4 times the third row from the second to finally get the row reduced echelon form as follows: \[\begin{bmatrix} 1 & 0 & 8/3 & 1/3 \ 0 & 1 & -2/3 & 2/3 \ 0 & 0 & 1 & 4 \end{bmatrix}\]

Step 4: Writing the solution

From the augmented matrix in its final form, we get \(x_1 = 1/3\), \(x_2 = 2/3\) and \(x_3 = 4\) as the solution to the system of equations.

Key Concepts

Understanding System of Linear Equations

A system of linear equations is a collection of one or more equations with the same set of unknowns. In our exercise, starting with a straightforward example, we have three equations involving variables \(x_1, x_2,\) and \(x_3\). Each equation provides unique information, and together they establish constraints that help pinpoint values for these variables.

To solve such systems, we aim to find the values for the unknowns that satisfy all simultaneous equations. The structure of these equations is linear, meaning the exponents of the variables are all 1. This linearity not only simplifies calculations but also makes graphical representations clearer, often represented as straight lines or, in higher dimensions, planes.

• Linear equations are typically written in the form \(ax + by + cz = d\).
• Each equation represents a plane in a 3D space. The solution is the point where all planes intersect.

In solving these, the object is to reduce the number of equations until each variable independently equals a definitive number.

Matrix Equation and Its Representation

A matrix equation such as \(AX = B\) succinctly represents a system of linear equations in matrix form. In this format:

- \(A\) is a matrix that captures the coefficients of the variables.- \(X\) is a column matrix representing the variables.- \(B\) is a column matrix of constant terms on the right side of the equations.

This transformation from equations to matrices brings several advantages, including the ability to utilize powerful computational tools and techniques. The matrix format is particularly useful because it aligns well with systematic computational methods like Gauss-Jordan elimination, which is a method used to solve such equations.

• The matrix equation makes it easier to manipulate multiples equations at once.
• Operations on matrices reflect equivalent operations on the equations when solving them.

This method streamlines the solving process by simplifying complex equations into a more manageable form.

Achieving Row Reduced Echelon Form with Gauss-Jordan Elimination

Row reduced echelon form (RREF) is a specific type of matrix form achieved using the Gauss-Jordan elimination method. The aim of bringing a matrix to this form is to simplify it to the point where the solution to the system of linear equations becomes obvious.

The Gauss-Jordan elimination involves a series of row operations on the augmented matrix \([A | B]\) to transform it into a form where

• The first non-zero number in each row (the "leading 1") is 1.
• Each leading 1 is the only non-zero number in its column.
• Leading 1s move from left to right going downward.
• When there are no leading 1s, they are placed all the way to the right, giving zeros below and above them.

Once a matrix is in this form, it clearly shows the values of variables directly, making them easy to read off from the matrix: \(x_1 = 1/3\), \(x_2 = 2/3\), and \(x_3 = 4\) in our specific exercise case. This process effectively uncovers the solution hidden within the seemingly complex web of the original systems of equations.