Problem 17 - College Algebra

Question

In Exercises \(13-18\), write the system of linear equations represented by the augmented matrix. Use \(x, y, z\) and, if necessary, \(w, x, y,\) and \(z,\) for the variables. Once the system is written, use back substitution to find its solution. \(\left[\begin{array}{rrrr|r}1 & -1 & 1 & 1 & 3 \\ 0 & 1 & -2 & -1 & 0 \\ 0 & 0 & 1 & 6 & 17 \\ 0 & 0 & 0 & 1 & 3\end{array}\right]\)

Step-by-Step Solution

Answer

The solutions to the system of equations are \(x=2\), \(y=1\), \(z=-1\), and \(w=3\).

Step 1: Write the System of Equations

First, write the system of the linear equations from the augmented matrix. Each row of the matrix correspond to an equation. The augmented matrix \[ \left[\begin{array}{rrrr|r}1 & -1 & 1 & 1 & 3 \ 0 & 1 & -2 & -1 & 0 \ 0 & 0 & 1 & 6 & 17 \ 0 & 0 & 0 & 1 & 3\end{array}\right] \] represents four equations: \(x-y+z+w=3\), \(y-2z-w=0\), \(z+6w=17\), \(w=3\).

Step 2: Substitute from Bottom to Top

Starting from the bottom of the system of equations, we first find that \(w=3\). We substitute \(w=3\) into equation for \(z\), giving \(z+6*(3)=17\), so that \(z=-1\). Next we substitute \(w=3\) and \(z=-1\) into equation for \(y\), giving \(y-2*(-1)-3=0\), so that \(y=1\). Finally substitute \(w=3\), \(z=-1\), and \(y=1\) into equation for \(x\), giving \(x-1+(-1)+3=3\), so that \(x=2\).

Step 3: Check The Solution

Finally for a last step, we plug the obtained values of \(x, y, z, w\) back into the system of equations to ensure the solutions are correct. Plugging back the answers into each equation, we verify that indeed all of them hold true.

Key Concepts

System of Linear Equations

When dealing with a system of linear equations, we're essentially looking at multiple linear equations that we aim to solve simultaneously. The solution to this system is the set of values that satisfy all equations at once. In the context of our exercise, the augmented matrix provides a condensed way to represent these equations.

The given matrix corresponds to a set of equations where each row represents an equation and the columns up to the last represent the coefficients of the variables. The rightmost column stands for the constant terms. For example, the first row represents the equation \(x-y+z+w=3\). Here, the numbers 1, -1, 1, and 1 are the coefficients of \(x\), \(y\), \(z\), and \(w\) respectively, with 3 being the constant term. To solve these equations, we can use methods like Gaussian elimination, which is why they are arranged in this particular format.

The goal is to find a common solution for \(x\), \(y\), \(z\), and \(w\) that makes all four equations true simultaneously. This set of numbers is what we call the solution of the system of linear equations.

Back Substitution

The concept of back substitution comes into play after you have reduced your system of equations into an upper triangular form, as represented by the augmented matrix. This is a standard technique in linear algebra to solve systems, particularly useful when dealing with several variables.

Once we've arranged our equations, we start from the last one, which contains only one variable, and work our way up. For instance, the bottom row of our matrix gives \(w=3\) directly. With this value of \(w\), we substitute it back into the equation above it to find the value of \(z\). Continuing this process, we systematically solve for each variable.

Why is it effective?

Back substitution is effective because it simplifies each equation to a single variable, removing the need to solve the equations simultaneously. This process harnesses the concept of dependency among the variables, as each one found can be used to determine the others, one step at a time.

Algebraic Substitution

On the other hand, algebraic substitution is an essential technique where we replace one variable with another equivalent value based on a different equation. It is primarily used when two equations are interdependent i.e., one contains a variable that can be defined in terms of another from a different equation.

While back substitution works downward from the last equation, algebraic substitution does not have a strict order; it depends on the equations at hand. For instance, if you have an equation where \(y=2x+1\) and another equation involving \(y\), you can 'substitute' the expression for \(y\) from the first equation into the second. This allows you to express the second equation solely in terms of \(x\), making it easier to solve.

Practical Application

Once you're comfortable with both back and algebraic substitution, you can apply these techniques to any set of linear equations, be they part of your mathematics coursework or even complex real-world problems involving linear relationships.