Problem 54 - College Algebra and Calculus: An Applied Approach

Question

An augmented matrix that represents a system of linear equations (in variables \(x, y\), and \(z\) ) has been reduced using Gauss-Jordan elimination. Write the solution represented by the augmented matrix. $$ \left[\begin{array}{rrrrr} 1 & 0 & 2 & \vdots & 9 \\ 0 & 1 & 5 & \vdots & -3 \\ 0 & 0 & 0 & \vdots & 0 \end{array}\right] $$

Step-by-Step Solution

Answer

The solution to the system of equations represented by the given augmented matrix is \(x = 9 - 2t\), \(y = -3 - 5t\), and \(z = t\), where \(t\) can be any real number.

Step 1: Translate matrix to system of equations

The first step is to convert the matrix back into a system of linear equations. Each row of the matrix represents one equation. The first column gives the coefficient of \(x\), the second the coefficient of \(y\), the third gives the coefficient of \(z\) and the number in the last column (on the right side of the bar) is the constant. Given matrix is interpreted as follows:\n\n\(x + 2z = 9\)\n+\(y + 5z = -3\)\n0 'equation' translates into \(0 = 0\)

Step 2: Identify solutions

Second step is to identify the solutions to the linear equations. The third equation does not provide any value as it translates into \(0 = 0\). This means that \(z\) is a free variable which can take any real number. From the first equation, we get \(x = 9 - 2z\), and the second equation yields, \(y = -3 - 5z\).

Step 3: Write solution in parameterized form

Finally, the solution is written in the form where \(z\) can be any real number, usually in terms of parameter \(t\). So, substitute \(z = t\), then the parameterized solution becomes \(x = 9 - 2t\), \(y = -3 - 5t\), and \(z = t\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. In the context of the exercise, you are dealing with three variables: \(x, y,\) and \(z\). The goal is to find the values of these variables that satisfy all the equations simultaneously.
When you look at a system, each equation is a straight line in the multidimensional space. The solution to the system is the point or points where these lines intersect. This exercise deals with two equations after a reduction process, which appear as follows:

• \(x + 2z = 9\)
• \(y + 5z = -3\)

The process of solving such systems often involves simplifying these equations using techniques like Gaussian elimination or Gauss-Jordan elimination. Reduction techniques aim to bring the system into a simpler equivalent form where the solution is more apparent. The simplified form in this problem shows that \(z\) has no specific value, making it a free variable. Instead, we express \(x\) and \(y\) in terms of \(z\).

Augmented Matrix

An augmented matrix is a crucial concept that helps in handling systems of linear equations in a structured form. It combines the coefficients of the variables and the constants from each equation into a single matrix. This matrix makes it easier to perform operations and apply elimination methods.
To form an augmented matrix, you take all the coefficients from each equation's left side and place them as columns of a matrix. You then add a vertical line (often represented as \(\vdots\)) to separate these from the constants.
In the given exercise, the augmented matrix represents the system:

• \(1x + 0y + 2z = 9\)
• \(0x + 1y + 5z = -3\)
• \(0x + 0y + 0z = 0\)

The augmented matrix, in this case, is:\[\begin{bmatrix}1 & 0 & 2 & \vdots & 9 \0 & 1 & 5 & \vdots & -3 \0 & 0 & 0 & \vdots & 0 \\end{bmatrix}\]This representation makes it very efficient to apply methods like Gauss-Jordan elimination to find the solutions.

Matrix Solution

Finding the solution to a system of linear equations using matrices involves several steps that eventually allow for understanding the relationships between variables. After reducing a matrix using the Gauss-Jordan elimination method, you're often left with a simpler system which can be directly explained or solved further using parameters.
In this exercise, after the reduction, you have two significant linear equations and one trivial one: \(0 = 0\). The important aspect of this outcome is that \(z\) becomes a free variable, signifying it can adopt any value. With \(z\) not fixed, both \(x\) and \(y\) are expressed in terms of \(z\):

• \(x = 9 - 2z\)
• \(y = -3 - 5z\)
• \(z = t\) where \(t\) is any real number

This parameterized approach implies solutions are not unique and can form a line in three-dimensional space, making the solution more flexible and offering infinite possibilities. This conclusion is possible due to the existence of a free variable, which in many real-world applications, gives a more adaptable system of equations.