Problem 48 - College Algebra and Calculus: An Applied Approach

Question

(a) write the system of linear equations as a matrix equation \(A X=B\), and (b) use Gauss-Jordan elimination on the augmented matrix \([A: B]\) to solve for the matrix \(X\). $$ \left\\{\begin{aligned} 2 x-4 y+z &=0 \\ -x+3 y+z &=1 \\ x+y &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions to the system of equations are x = 2, y = 1, z = 0.

Step 1: Represent the system in Matrix Form (AX = B)

Identify coefficients matrix (A), variable matrix (X) and constant matrix (B) to frame the matrix equation. Matrix A : \(\begin{bmatrix} 2 & -4 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 0 \end{bmatrix}\), Matrix X : \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\), Matrix B : \(\begin{bmatrix} 0 \\ 1 \\ 3 \end{bmatrix}\). Thus, the matrix equation is \(AX = B\) which is \(\begin{bmatrix} 2 & -4 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 3 \end{bmatrix}\)

Step 2: Augmented matrix formation

Combine matrices A and B as an augmented matrix \([A : B]\). It will be \(\begin{bmatrix} 2 & -4 & 1 & : & 0 \\ -1 & 3 & 1 & : & 1 \\ 1 & 1 & 0 & : & 3 \end{bmatrix}\)

Step 3: Apply Gauss-Jordan elimination

Transform the augmented matrix into reduced row echelon form (rref). First, swap row 1 with row 3. After this operation, the augmented matrix becomes \(\begin{bmatrix} 1 & 1 & 0 & : & 3 \\ -1 & 3 & 1 & : & 1 \\ 2 & -4 & 1 & : & 0 \end{bmatrix}\). Next, add row 1 to row 2 and multiply row 2 by 2. The augmented matrix transforms to \(\begin{bmatrix} 1 & 1 & 0 & : & 3 \\ 0 & 4 & 1 & : & 4 \\ 2 & -4 & 1 & : & 0 \end{bmatrix}\). Lastly, subtract 2 times row 1 from row 3 and divide row 2 by 4 to get \(\begin{bmatrix} 1 & 1 & 0 & : & 3 \\ 0 & 1 & 0.25 & : & 1 \\ 0 & -6 & 1 & : & -6 \end{bmatrix}\). Adding 6 times row 2 to row 3 we get the final form \(\begin{bmatrix} 1 & 0 & 0 & : & 2 \\ 0 & 1 & 0.25 & : & 1 \\ 0 & 0 & 2.5 & : & 0 \end{bmatrix}\).

Step 4: Solve for Variables

The elements on the left side of the colon form the identity matrix, and the elements on the right side represent variables x, y, z respectively. Solving the final equations, we have \(x = 2, y = 1, z = 0\).

Key Concepts

Matrix Equation

A matrix equation is a way to compactly represent a system of linear equations. In the exercise, a set of three linear equations is presented. These equations involve variables \(x\), \(y\), and \(z\). The key is to express these relationships in matrix form, using a coefficient matrix \(A\), a variable matrix \(X\), and a constant matrix \(B\). The matrix equation takes the form \(AX = B\), where:

• \(A\) is the matrix of coefficients from the left side of the equations.
• \(X\) is the column matrix containing the unknowns \(x, y,\) and \(z\).
• \(B\) is the column matrix of constants from the right side of the equations.

For this exercise: \[ A = \begin{bmatrix} 2 & -4 & 1 \ -1 & 3 & 1 \ 1 & 1 & 0 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} 0 \ 1 \ 3 \end{bmatrix} \] The matrix equation is \[\begin{bmatrix} 2 & -4 & 1 \ -1 & 3 & 1 \ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 1 \ 3 \end{bmatrix}\] This structure simplifies operations when solving the system.

Augmented Matrix

An augmented matrix is an essential tool in solving systems of linear equations. It arises by merging the coefficient matrix \(A\) and the constant matrix \(B\) into one matrix, separated by a colon or a division line. Essentially, this matrix includes all the information contained in the initial system of equations:

• The augmented matrix for our system is written as: \[\begin{bmatrix} 2 & -4 & 1 & : & 0 \ -1 & 3 & 1 & : & 1 \ 1 & 1 & 0 & : & 3 \end{bmatrix}\]
• The left side of the colon consists of the coefficients of the variables.
• The right side holds the constants from the equations.

The benefit of using an augmented matrix is that it provides a framework for systematically applying row operations, which will help simplify the matrix and ultimately solve the system of equations using methods like Gauss-Jordan elimination.

Reduced Row Echelon Form

The reduced row echelon form (RREF) is a specific form of a matrix that is achieved by applying row operations. The goal is to simplify the augmented matrix so that it directly reveals the solutions to the system of linear equations. In the Gauss-Jordan elimination method, particular conditions define RREF:

• Each leading entry in a row is 1, and it is the only non-zero entry in its column.
• The leading 1 of each row appears to the right of the leading 1s in the rows above it.
• Rows consisting entirely of zeros, if any, are at the bottom of the matrix.

In the solution:\[\begin{bmatrix} 1 & 0 & 0 & : & 2 \ 0 & 1 & 0.25 & : & 1 \ 0 & 0 & 2.5 & : & 0 \end{bmatrix}\]This matrix tells us the values of the variables directly, since:- The first row implies \(x = 2\)- The second row implies \(y = 1\)- And the adjusted third row gives \(z = 0\).

Linear Equations

Linear equations are equations of the first order, involving variables raised to the power of one. They are called linear because they graph as straight lines. In the exercise, we deal with three such equations:

• \(2x - 4y + z = 0\)
• \(-x + 3y + z = 1\)
• \(x + y = 3\)

The primary goal is to find values of \(x\), \(y\), and \(z\) that satisfy all the equations concurrently. Mathematically, this is akin to finding the point of intersection of the lines or planes represented by these equations. Linear equations are straightforward and can be solved by several methods:

• Substitution and elimination over smaller sets.
• Matrix operations like those demonstrated here for larger systems.

In systems like this one, Gauss-Jordan elimination with matrices is highly efficient for finding solutions.