Problem 16 - College Algebra

Question

Solve the system by Gaussian elimination. \(\left[\begin{array}{ll|l}1 & 0 & 3 \\ 0 & 0 & 0\end{array}\right]\)

Step-by-Step Solution

Answer

The solution is \( x = 3 \), with no other variables involved.

Step 1: Write the System from the Augmented Matrix

The given augmented matrix represents the system of linear equations. The matrix \( \begin{bmatrix} 1 & 0 & | & 3 \ 0 & 0 & | & 0 \end{bmatrix} \) implies the following system of equations: 1. \( x = 3 \) 2. \( 0 = 0 \). In mathematical terms, the second equation is always true and does not provide additional constraints.

Step 2: Interpret the System

Since the first equation simplifies to \( x = 3 \) and the second equation is \( 0 = 0 \), we find that the system is consistent. The second row does not place any restrictions on variables, hence it suggests there are either infinitely many solutions or a specific solution. Here, given only one equation effectively, \( x \) is isolated and already determined.

Step 3: Analyze the Resulting Solution

With the system providing \( x = 3 \), there are no other variables involved or free variables to consider. Since the row for the second variable was zero, the solution is complete as far as \( x \) is concerned. We conclude that there is a single solution for this system, specifically related to the variable \( x \).

Key Concepts

Augmented Matrix

An augmented matrix is an essential tool in solving linear equations. It combines both the coefficients of the variables and the constants from the equations into a single matrix form. This streamlined form helps simplify operations like Gaussian elimination. The structure of an augmented matrix includes all the information needed to solve a system of equations by using row operations.

Consider the augmented matrix given in the problem: \[ \left[\begin{array}{cc|c}1 & 0 & 3 \ 0 & 0 & 0\end{array}\right] \]This represents a set of linear equations where each row corresponds to an equation. The vertical line in the matrix separates the coefficients of the variables from the constants. When expressed in this form, it becomes straightforward to manipulate and solve the equations using techniques like Gaussian elimination.

System of Equations

A system of equations consists of multiple equations that share the same set of variables. Solving such a system means finding the values of these variables that satisfy all the equations at once. In the given exercise, the system derived from the augmented matrix translates to:

• \( x = 3 \)
• \( 0 = 0 \), a trivial truth that holds for any \( x \)

This form reveals a simplified system where each equation aligns with a row from the augmented matrix. Solving these systems can involve strategies like substitution or elimination, or using matrices for more complex cases. The primary goal is to determine whether solutions exist. This often leads to identifying whether the system is consistent or inconsistent.

Consistent System

A consistent system in the context of linear equations is one where at least one set of solutions exists that satisfies every equation in the system. The opposite, an inconsistent system, would have no solutions. A crucial step in solving using matrices is determining the system's consistency.

In our exercise, since the only non-trivial equation is \( x = 3 \), and the remaining equation is a tautology \( 0 = 0 \), we conclude that the system is consistent. Consistency indicates the absence of contradictions in the system of equations. This means, in this context, there exists a solution \( x = 3 \) that does not contradict the other 'equation.' The row of zeros adds no restrictions, reinforcing consistency without further constraints.

Infinite Solutions

Infinite solutions occur in a system of equations when there are more variables than constraints, leading to many values satisfying the equations. Typically, systems with infinite solutions arise when rows of zeros appear in the augmented matrix without leading to contradictions.

In the given problem, however, although a row of zeros is present, it does not mean there are infinite solutions. Instead, the matrix efficiently reduces to a scenario with a determined solution for \( x \). The equation \( 0 = 0 \) does not introduce any new variables or need for adjustments, leading to a unique conclusion. In cases where more variables had existed and remained unbounded by rigorous equations, infinite solutions would be a valid result. Here, nonetheless, the system's closure on \( x \) alone halts any such expansion towards multiple solutions.