Problem 14 - College Algebra

Question

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\begin{array}{rr}3 w+2 x-y+2 z= & -12 \\\4 w-x+y+2 z= & 1 \\\w+x+y+z= & -2 \\\\-2 w+3 x+2 y-3 z= & 10\end{array}$$

Step-by-Step Solution

Answer

The complete solution to the given system is \(w = -2\), \(x = z/3\), \(y = -2x/3\) and \(z = 56/5\).

Step 1: Writing the Augmented Matrix

The equation system can be converted into the equivalent augmented matrix: \[\begin{bmatrix} 3 & 2 & -1 & 2 & | & -12 \\4 & -1 & 1 & 2 & | & 1 \\1 & 1 & 1 & 1 & | & -2 \\ -2 & 3 & 2 & -3 & | & 10 \\ \end{bmatrix}\]

Step 2: Forming A Triangular Matrix

Then, perform the elementary row operations. Swap first and third row. Then, multiply first row by 3, second by 4 and fourth by 1 for convenience and reducing the below rows to form a triangular matrix: \[\begin{bmatrix} 3 & 3 & 3 & 3 & | & -6 \\0 & -5 & -5 & -2 & | & 8 \\0 & -3 & -14 & -6 & | & 32 \\0 & 11 & 8 & -9 & | & 16 \\ \end{bmatrix}\]

Step 3: Row Reduction to Echelon Form

Keep on performing the elementary row operations to reduce the matrix to what’s known as row-echelon form: \[\begin{bmatrix} 3 & 3 & 3 & 3 & | & -6 \\0 & 1 & 1 & 2/5 & | & -8/5 \\0 & 0 & -1 & 3/-10 & | & 52/5 \\0 & 0 & 1 & -16/5 & | & -56/5 \\ \end{bmatrix}\]

Step 4: Back Substitution to solve for variables

Lastly, perform back substitution to find the variables. The system becomes following as per our last matrix: \(w = -2 - x - y - z\), \(x = -8/5 -y -2z/5\), \(y = -52/5 +3z/10\), and \(z = 56/5\). Substituting z into y we get \(y = -2x/3\). Then subs in x equation \(x = -8/5 - (-2x/3) -2z/5\) we find \(x = z/3\). Lastly, \(w = -2 - z/3+ 2z/3 - z = -2\).

Key Concepts

Augmented Matrix

To tackle a system of equations using Gaussian elimination, the first step is creating an augmented matrix. This matrix includes all coefficients and constants from the system of equations.
The left side of the vertical bar contains the coefficients of the variables, while the right side shows the constant terms. This format helps in keeping track of solutions as we use various operations.
For example, given a system of equations, the augmented matrix might look like this:

• The equation coefficients from the system translate to rows in the matrix.
• Each column, except the last, represents a different variable in the system.
• The last column after the vertical bar holds the constants.

This setup helps streamline the process before beginning row operations, setting the stage for the next transformation to the row-echelon form.

Row Echelon Form

Turning an augmented matrix into row-echelon form simplifies solving systems of linear equations.
In this form, each row after the first starts with a zero and we aim for a stair-step pattern. This transformation uses elementary row operations like swapping rows, multiplying rows by a non-zero constant, or adding multiples of one row to another.
The key goal here is to create leading 1s, or pivots, in each row moving diagonally down the matrix. This stair-step arrangement facilitates easier back substitution later.
Ultimately, achieving a row-echelon form indicates the system is ready for the final stages of solving as it reveals the interdependencies between variables clearly.

Back Substitution

Once the matrix is in row-echelon form, we use back substitution to find solutions for the variables.
Starting with the bottom-most row, where typically only one or two variables are present, we solve for these variables directly from the equations.
Then, we substitute these values back into equations represented by higher rows, unraveling more variables step-by-step.
This process ensures that we utilize the simplified linear relationships established by the row-echelon form, systematically deriving values for each variable until all are determined.

Elementary Row Operations

Elementary row operations are the three basic manipulations we perform on the rows of an augmented matrix to simplify it:

• Row Switching: Swapping two rows helps if a row beneath has a better pivot.
• Row Multiplication: Multiplying any row by a nonzero scalar helps in normalizing pivots.
• Row Addition: Adding a multiple of one row to another helps eliminate coefficients aiming for zeros.

Using these operations correctly can transform a matrix significantly without changing the system's solutions.
This is crucial to achieving both row-echelon and reduced row-echelon forms during Gaussian elimination, enabling a straightforward path to find solutions through back substitution.