Problem 22 - Intermediate Algebra

Question

For Problems \(1-22\), solve each of the systems and use matrices as we did in the examples of this section. $$ \left(\begin{array}{rr} 2 x+5 y+z= & 1 \\ x+2 y-3 z= & -13 \\ 3 x-y-2 z= & -4 \end{array}\right) $$

Step-by-Step Solution

Answer

The solutions are \( x \approx 3.21 \), \( y \approx -0.79 \), \( z \approx -3.57 \).

Step 1: Write the System in Matrix Form

We start by rewriting the given system of equations in matrix form. The system \( \begin{align*} 2x + 5y + z &= 1 \ x + 2y - 3z &= -13 \ 3x - y - 2z &= -4 \end{align*} \) can be expressed in the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the constant matrix. Thus, \( A = \begin{bmatrix} 2 & 5 & 1 \ 1 & 2 & -3 \ 3 & -1 & -2 \end{bmatrix} \), \( X = \begin{bmatrix} x \ y \ z \end{bmatrix} \), and \( B = \begin{bmatrix} 1 \ -13 \ -4 \end{bmatrix} \).

Step 2: Form the Augmented Matrix

Next, create the augmented matrix \([A|B]\) by combining matrix \( A \) and matrix \( B \). Thus, \([A|B] = \begin{bmatrix} 2 & 5 & 1 & | & 1 \ 1 & 2 & -3 & | & -13 \ 3 & -1 & -2 & | & -4 \end{bmatrix}\).

Step 3: Apply Row Operations

Use row operations to transform the augmented matrix into row-echelon form. Start by making the first column below the main diagonal zero. Convert \( R2 \) by subtracting \( 0.5 \times R1 \) from \( R2 \). Then, convert \( R3 \) by subtracting \( 1.5 \times R1 \) from \( R3 \). The matrix becomes \( \begin{bmatrix} 2 & 5 & 1 & | & 1 \ 0 & -0.5 & -3.5 & | & -13.5 \ 0 & -8.5 & -3.5 & | & -5.5 \end{bmatrix} \).

Step 4: Continue Row Operations for Simplification

Make the second column below the main diagonal zero. Use \( R3 = R3 - 17 \times R2 \) to simplify. This transforms the matrix to \( \begin{bmatrix} 2 & 5 & 1 & | & 1 \ 0 & -0.5 & -3.5 & | & -13.5 \ 0 & 0 & -63 & | & 224.5 \end{bmatrix} \).

Step 5: Back Substitution

Once the matrix is in row-echelon form, solve the system using back substitution. Starting with the last row, solve for \( z \) which gives \( z = -3.57 \) (approximate). Using this, substitute back into the second row to solve for \( y \), and then into the first row for \( x \). The solutions are approximate: \( x \approx 3.21 \), \( y \approx -0.79 \), and \( z \approx -3.57 \).

Key Concepts

Linear System of Equations

A linear system of equations is a collection of two or more linear equations involving the same set of variables. In our given problem, we have a system of three equations involving three variables: \(x\), \(y\), and \(z\). This is expressed as: - \(2x + 5y + z = 1\) - \(x + 2y - 3z = -13\) - \(3x - y - 2z = -4\) Its main goal is to find values for \(x\), \(y\), and \(z\) that satisfy all equations simultaneously.To solve this system, we transform it into a matrix format which makes calculations easier and more systematic. Simultaneous solutions or intersecting points in geometric perspective are possible outcomes of such systems. Understanding and solving these systems are fundamental in fields like physics, engineering, and economics. Each equation represents a plane, and the solution to the system is the point where these planes intersect.

Row-Echelon Form

Row-echelon form is a type of matrix form used to simplify the process of solving a system of linear equations. A matrix in this form has a staircase-like structure and fulfills the following conditions: - Each non-zero row begins with a leading 1, which is positioned further to the right than the leading 1 in any previous row. - The leading entries below the main diagonal are all zeros. - Zero rows, if any, are at the bottom of the matrix. You achieve this form through row operations, which involve swapping rows, scaling rows, or adding/subtracting rows from each other. In the exercise, transforming the augmented matrix to row-echelon form allows us to use back substitution efficiently, starting from the bottom row to solve for variables step-by-step.

Augmented Matrix

An augmented matrix is an essential tool in solving linear systems. It combines the coefficient matrix and the constant matrix into one. By doing this, it visually aligns the system of equations into a convenient form for applying matrix operations.For the given problem, the augmented matrix is written as: \[\begin{bmatrix} 2 & 5 & 1 & | & 1 \ 1 & 2 & -3 & | & -13 \ 3 & -1 & -2 & | & -4 \end{bmatrix}\]The vertical line separates the coefficients of variables from their respective constants. The manipulation of this matrix using row operations leads it towards row-echelon form and ultimately towards solutions. By working with augmented matrices, one can handle both the variables and their equations simultaneously, simplifying complex systems into more manageable steps.