Problem 74 - Precalculus with Limits

Question

In Exercises 63-84, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. \( \left\\{ \begin{array}{l} 2x + 2y - z = 2 \\ x - 3y + z = -28 \\ -x + y = 14 \end{array} \right. \)

Step-by-Step Solution

Answer

The solution to the system of equations is \( x=-1 \), \( y=7 \) and \( z=-12 \)

Step 1: Convert System of Equations to Matrix

Transform the three equations into an augmented matrix. This gives us: \[ \left[ \begin{array}{ccc|c} 2 & 2 & -1 & 2 \ 1 & -3 & 1 & -28 \ -1 & 1 & 0 & 14 \end{array} \right] \]

Step 2: Apply Gaussian Elimination to Create Upper Triangular Matrix

Swap the 2nd row with the 1st one and multiply it by 1/2. Then add 2nd row to the 3rd row. The obtained upper triangular matrix will be: \[ \left[ \begin{array}{ccc|c} 1 & -3 & 1 & -28 \ 2 & 2 & -1 & 2 \ 0 & 0 & 1 & -12 \end{array} \right] \]

Step 3: Apply Back-Substitution to Extract Results

The second row becomes, from subtraction of two times the first row: \[ 2 - 2 = 0, 2 - (-6) = 8, -1 - 2 = -3, 2 - (-56) = 58 \\ The matrix becomes:\[ \left[ \begin{array}{ccc|c} 1 & -3 & 1 & -28 \ 0 & 8 & -3 & 58 \ 0 & 0 & 1 & -12 \end{array} \right] \] From here we can see that \( z = -12 \). Substitute \( z = -12 \) into the second equation, we get \( y = 7 \), and substitute \( y = 7 \) and \( z = -12 \) into the first equation, we get \( x = -1 \).

Step 4: Verify the solution

The solutions can be verified by plugging them back into the original equations. They hold true for all equations, therefore the solution \( x=-1, y=7, z=-12 \) is verified.

Key Concepts

Systems of Equations

A system of equations consists of multiple equations that share the same set of variables. These equations are solved together, allowing us to find a common solution that satisfies all of them simultaneously.
In the context of linear systems, which involve linear equations, these can be expressed in the form:

• The simplest form: two variables, two equations
• More complex forms: three variables and equations, like in our example

The main goal is to find the values for the variables that make every equation in the system true. This can be achieved through various methods, one of the most common is Gaussian elimination, which systematically reduces the equations to make them easier to handle.

Augmented Matrix

An augmented matrix is a simplified representation of a system of linear equations. It compacts the information of the coefficients and constants of each equation into a matrix format.
When forming an augmented matrix, you place each equation's coefficients in a row, stacking one row for each equation, and append a column for the constants on the right.
For example, the system of equations:

• \(2x + 2y - z = 2\)
• \(x - 3y + z = -28\)
• \(-x + y = 14\)

becomes: \[ \left[ \begin{array}{ccc|c} 2 & 2 & -1 & 2 \ 1 & -3 & 1 & -28 \ -1 & 1 & 0 & 14 \end{array} \right] \] This matrix layout provides an organized framework for the application of elimination techniques like Gaussian elimination.

Back-Substitution

Back-substitution is a method used to solve a system of linear equations once the equations are converted into an upper triangular matrix form via Gaussian elimination.
In an upper triangular matrix, you can solve for variables starting from the last row (since it will have only one variable) and continue upwards.
For instance, if the matrix yields: \[ \left[ \begin{array}{ccc|c} 1 & -3 & 1 & -28 \ 0 & 8 & -3 & 58 \ 0 & 0 & 1 & -12 \end{array} \right] \]We can see directly that:\( z = -12 \). Then substitute \( z \) back into the second row to find \( y \), and so on, until all variables are found.
Back-substitution simplifies complex systems, turning them into a straightforward problem-solving exercise from the bottom row up.

Upper Triangular Matrix

An upper triangular matrix is a type of matrix where all the entries below the main diagonal are zeros. This structure allows for straightforward backward solving, as each row can be interpreted as an equation with decreasing complexity as you move up the matrix.
The goal in Gaussian elimination is to transform the original coefficient matrix into an upper triangular form. This often involves:

• Swapping rows
• Multiplying rows by constants
• Adding or subtracting multiples of one row to another

In our example, the transformation that led to the matrix being upper triangular involved the operations to zero out the lower half of the triangle.
Once in this form, solving the system is significantly more straightforward since each step of back-substitution uses only the known values from previous steps. This process greatly simplifies the task of solving a system of equations.