Problem 27 - Precalculus: Mathematics for Calculus

Question

A matrix is given in row-echelon form. (a) Write the system of equations for which the given matrix is the augmented matrix. (b) Use back-substitution to solve the system. $$\left[\begin{array}{rrrrr} 1 & 2 & 3 & -1 & 7 \\ 0 & 1 & -2 & 0 & 5 \\ 0 & 0 & 1 & 2 & 5 \\ 0 & 0 & 0 & 1 & 3 \end{array}\right]$$

Step-by-Step Solution

Answer

(7, 3, -1, 3).

Step 1: Identify the Variables

The given matrix corresponds to a system of equations. Each row represents an equation, and each column (excluding the last column, which is the constants) represents a variable. Let the variables be \(x_1, x_2, x_3,\) and \(x_4\).

Step 2: Write the System of Equations

Using the coefficients from each row in the matrix, write down the corresponding system of equations:1. The first row gives: \(x_1 + 2x_2 + 3x_3 - x_4 = 7\)2. The second row gives: \(x_2 - 2x_3 = 5\)3. The third row gives: \(x_3 + 2x_4 = 5\)4. The fourth row gives: \(x_4 = 3\).

Step 3: Back-Substitution - Solve for x4

From the fourth equation \(x_4 = 3\), we directly determine the value of \(x_4\).

Step 4: Back-Substitution - Solve for x3

Using the value \(x_4 = 3\) in the third equation, solve for \(x_3\):\(x_3 + 2(3) = 5\). Thus, \(x_3 = 5 - 6 = -1\).

Step 5: Back-Substitution - Solve for x2

Use the value \(x_3 = -1\) in the second equation to solve for \(x_2\):\(x_2 - 2(-1) = 5\).This simplifies to \(x_2 + 2 = 5\), so \(x_2 = 3\).

Step 6: Back-Substitution - Solve for x1

Substitute \(x_2 = 3, x_3 = -1, x_4 = 3\) into the first equation:\(x_1 + 2(3) + 3(-1) - 3 = 7\).Simplifying:\(x_1 + 6 - 3 - 3 = 7\)This reduces to \(x_1 = 7\).

Step 7: Solution

The solution to the system of equations is \(x_1 = 7\), \(x_2 = 3\), \(x_3 = -1\), and \(x_4 = 3\).

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with a shared set of unknowns. In solving these systems, we aim to find values for the variables that satisfy all equations simultaneously.
This particular problem utilizes four equations with four variables: \(x_1, x_2, x_3,\) and \(x_4\). These equations were crafted from a given augmented matrix in row-echelon form.
Here’s a few things to keep in mind about systems of equations:

• Each equation can provide useful information about the values of the variables when solved together.
• Methods like substitution, elimination, and matrix operations help find the solution to these equations.
• Having the same number of equations as variables usually means a solvable system—though the determinant should be non-zero.

In matrix terms, each row represents an equation, and the constants from the rightmost column are the results the equations equal. By finding solutions to these equations, we determine the values for each variable.

Row-Echelon Form

Row-echelon form is a way of organizing a matrix that simplifies the process of solving systems of linear equations. In this form, the matrix presents a staircase-like pattern when viewed from the left.
The matrix is structured such that each leading coefficient (the first non-zero number from the left in each row) is to the right of the leading coefficient in the row just above it. Furthermore, all entries below a leading coefficient are zeros.

• Helps in applying back-substitution to efficiently find solutions for variables.
• Provides an organized manner to visualize the steps needed to eliminate variables.

The matrix given in the exercise here is already in row-echelon form and consists of four rows, each corresponding to one of the equations in our system. Recognizing this form is pivotal as it directly allows us to apply the back-substitution method.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It's essentially a combination of the coefficient matrix and the constant terms in the equations. The vertical bar in an augmented matrix distinguishes between the coefficients of the unknowns and the constants from the equations.

• Simplifies the process of handling systems numerically by combining the information in one compact form.
• The matrix operations, like row echelon transformation, aid in reducing complex systems to simpler forms.

In our example, the augmented matrix allowed for a direct reading of each equation's coefficients and constants. This facilitated the systematic application of back-substitution once the matrix was in row-echelon form.