Problem 85 - Precalculus with Limits

Question

In Exercises 85-90, use the matrix capabilities of a graphing utility to reduce the augmented matrix corresponding to the system of equations, and solve the system. \( \left\\{ \begin{array}{l} 3x + 3y + 12z = 6 \\ x + y + 4z = 2 \\ 2x + 5y + 20z = 10 \\ -x + 2y + 8z = 4 \\ \end{array} \right. \)

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = -2t \), \( y = 2 - 4t \) and \( z = t \). Here, \( t \) is a parameter and can be any real number.

Step 1: Converting system of equations to matrix format

First, the given system of equations is converted into an augmented matrix. This is done by simply writing down the coefficients of the variables in each of the equations in the system.The resulting augmented matrix is:\[\left[ \begin{array}{ccc|c} 3 & 3 & 12 & 6 \ 1 & 1 & 4 & 2 \ 2 & 5 & 20 & 10 \ -1 & 2 & 8 & 4 \end{array} \right]\]

Step 2: Performing row operations

Next, use row operations to simplify the matrix into row-echelon form, i.e., getting upper triangular matrix. This involves replacing rows with their multiples and adding or subtracting rows. For this system, we can divide the first row by 3, and then replace the second row with (second row - first row), the third row with (third row - 2*first row) and the fourth row with (fourth row + first row). The matrix will now look like this:\[\left[ \begin{array}{ccc|c} 1 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \ 0 & 3 & 12 & 6 \ 0 & 3 & 12 & 6 \end{array} \right]\]

Step 3: Continuing row operations

Next, divide the third row by 3 and subtract the third row from the fourth row. The matrix will now look like this:\[\left[ \begin{array}{ccc|c} 1 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \ 0 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \end{array} \right]\]

Step 4: Translating Matrix into System of Equations and Solving

Now, translate the matrix back into a system of equations. The system will be as follows: \[\begin{cases} x + y + 4z = 2 \0 = 0 \y + 4z = 2 \0 = 0 \\end{cases}\]From these equations, we have two equations with three unknowns where the first and the third equations are the same. This system has infinitely many solutions which can be expressed in terms of parameter z: Let the free variable, z be t. Then from y + 4z = 2, we get y = 2 - 4t and from x + y + 4z = 2, we get x = -2t.

Key Concepts

Augmented Matrix

The concept of an augmented matrix is essential in solving systems of equations using matrices. Imagine taking all the coefficients from your system of equations and arranging them neatly in a grid-like structure. This grid includes not only the coefficients of your variables but also the constants from the equations themselves, separated by a vertical line.

• The left side represents the coefficients of the variables (like \(x\), \(y\), and \(z\)).
• The right side displays the constants from each equation.

For instance, considering our system of equations, the augmented matrix looks like this: ewline \[ \left[ \begin{array}{ccc|c} 3 & 3 & 12 & 6 \ 1 & 1 & 4 & 2 \ 2 & 5 & 20 & 10 \ -1 & 2 & 8 & 4 \end{array} \right] \] This matrix is a compact way to represent the entire system of equations and forms the basis for applying various techniques to find solutions.

Row Operations

Row operations are the secret sauce for simplifying augmented matrices into a form from which solutions can be easily obtained. These operations include:

• Swapping two rows.
• Multiplying a row by a nonzero constant.
• Adding or subtracting rows.

These operations help to transform the matrix into what's known as row-echelon form or even further into reduced row-echelon form. The goal here is to have zeros below the leading coefficients (the first nonzero number from the left) in each row. This simplifies solving the system. In our exercise, careful application of row operations leads to: \[ \left[ \begin{array}{ccc|c} 1 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \ 0 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \end{array} \right] \] By simplifying the matrix in such a structured manner, it becomes much easier to interpret and solve the system of equations.

Infinitely Many Solutions

A system of equations can have a unique solution, no solution, or infinitely many solutions. In our case, the matrix simplification sheds light on having infinitely many solutions.
When a system shows rows of zeros like the second and fourth rows in our transformed matrix: \[ \left[ \begin{array}{ccc|c} 1 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \ 0 & 1 & 4 & 2 \ 0 & 0 & 0 & 0 \end{array} \right] \] it suggests that there are not enough independent equations to pin down exactly one solution. Instead, the solution can vary based on a parameter. Here, \(z\) is chosen as a free variable named \(t\).
This gives the solutions expressible in terms of \(t\) as \(y = 2 - 4t\) and \(x = -2t\). Consequently, you'll find a different solution for each distinct value you assign to \(t\), leading to an infinite set of solutions.