Problem 11 - College Algebra

Question

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

Step-by-Step Solution

Answer

The complete solution to the system of equations is a one-parameter family of solutions given by \((w,x,y,z) = (24+8z, 1/11z + 53/11, -2z - 8, z)\). There exist infinitely many solutions because we have a free variable \(z\). For each value of \(z\), we will get a different solution set \((w,x,y,z)\).

Step 1: Form the Augmented Matrix

First, the system of equations is written as an augmented matrix. We set it up so that each row corresponds to an equation and each column corresponds to a coefficient of the variable in each equation. For the given equations, the augmented matrix will be \[\left[ \begin{array}{cccc|c} 2 & 1 & -1 & 0 & 3 \ 1 & -3 & 2 & 0 & -4 \ 3 & 1 & -3 & 1 & 1 \ 1 & 2 & -4 & -1 & -2 \end{array} \right]\].

Step 2: Perform Row Operations to Reach Row Echelon Form

Next, we want to use row operations to transform the matrix into row echelon form. We can do that in several steps: First, swap rows 1 and 2, then add row 2 multiplied by -2 to row 1, add row 2 multiplied by -3 to row 3 and finally add row 2 to row 4. The resulted matrix will be \[\left[ \begin{array}{cccc|c} 0 & 5 & -5 & 0 & 5 \ 1 & -3 & 2 & 0 & -4 \ 0 & 10 & -9 & 1 & 13 \ 0 & -1 & -2 & -1 & -6 \end{array} \right]\].

Step 3: Continue Row Operations to Reach Row Echelon Form

Subtract row 2 times 5 from row 1, subtract row 2 times 10 from row 3 and add row 2 to row 4. The resulting matrix is \[\left[ \begin{array}{cccc|c} 0 & 0 & 15 & 0 & 15 \ 1 & -3 & 2 & 0 & -4 \ 0 & 0 & 11 & 1 & 53 \ 1 & -2 & 0 & -1 & -10 \end{array} \right]\]. Now, divide row 1 by 15. Now, let's subtract row 2 from row 4 so we can also get a 0 under the 1 of the second column. After these operations, our matrix looks like this \[\left[ \begin{array}{cccc|c} 0 & 0 & 1 & 0 & 1 \ 1 & -3 & 2 & 0 & -4 \ 0 & 0 & 11 & 1 & 53 \ 0 & 1 & -2 & -1 & -6 \end{array} \right]\]. Now, we have to make the leading coefficients (or pivots) of each row be 1 and ensure that above and below each pivot is a column of zeros. We can achieve this by dividing row 3 by 11: \[\left[ \begin{array}{cccc|c} 0 & 0 & 1 & 0 & 1 \ 1 & -3 & 2 & 0 & -4 \ 0 & 0 & 1 & 1/11 & 53/11 \ 0 & 1 & -2 & -1 & -6 \end{array} \right]\]. Swap row 4 with row 2, subtract row 3 from row 1 and add row 3 times 2 to row 2: \[\left[ \begin{array}{cccc|c} 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & -2 & -8 \ 0 & 0 & 1 & 1/11 & 53/11 \ 1 & 0 & 0 & -8 & 24 \end{array} \right]\]. Swap row 1 with row 4, swap row 2 with row 3 and swap row 3 with row 4: \[\left[ \begin{array}{cccc|c} 1 & 0 & 0 & -8 & 24 \ 0 & 0 & 1 & 1/11 & 53/11 \ 0 & 1 & 0 & -2 & -8 \ 0 & 0 & 0 & 0 & 0 \end{array} \right]\]. The augmented matrix is now in row-echelon form.

Step 4: Interpret the Solution

Now we have an equivalent system of equations given by the last matrix. We can convert this back into equation form: \(w - 8z = 24\), \(y = -2z - 8\), and \(x =1/11z + 53/11\). Thus, we have a parametric solution: for any value of \(z\), we can find a corresponding solution for \(w\), \(x\), and \(y\). Thus, the solution is a one-parameter family of solutions given by \((w,x,y,z) = (24+8z, 1/11z + 53/11, -2z - 8, z)\). There exist infinitely many solutions because we have a free variable \(z\). For each value of \(z\), we will get a different solution set \((w,x,y,z)\). This is the complete solution of the system.

Key Concepts

Augmented Matrix

When we're given a system of linear equations, we can represent it as an augmented matrix to simplify the process of finding a solution. This matrix combines all the coefficients of the variables and the constant terms from each equation into one rectangular array. In an augmented matrix, each row corresponds to one equation and each column aligns with the coefficients of one of the variables.

For example, in our given system, the augmented matrix looks like this:

• Each row stands for one equation;
• Columns hold the coefficients of each variable and the last column holds the constants.

This arrangement makes it straightforward to apply Gaussian elimination, as we'll see with row operations.

Row Echelon Form

The row echelon form of a matrix is a simplified version achieved after applying a series of row operations. In this form:

• Leading coefficients (or pivots) of each row are 1.
• All entries below the pivots are zero.

To reach this form, we use elementary row operations like row swapping, scaling of rows, and adding/subtracting multiples of one row from another.

This step-by-step transformation makes the system easier to solve. Each operation performed on the augmented matrix corresponds to a valid manipulation of the system of equations. Once the matrix is in row echelon form, we can start back-substituting to find solutions.

Parametric Solution

A parametric solution is a way of expressing the solutions of a system of equations using parameters as variables. When an equation has more than one solution, especially infinitely many solutions, a parameter is used to represent the range of possible solutions.

For our system, once we reach row echelon form and back-substitute, we find expressions for each variable in terms of a free variable (in this case, \(z\)). These expressions show us the family of solutions depending on the value of \(z\). For example, we might express a solution like \(w = 24 + 8z\), where \(z\) can be any real number.

This flexibility allows us to understand that there is not just a single answer, but a whole set of answers depending on the free variable.

Free Variable

A free variable is a variable that is not bound by a pivot position in the row echelon form. It can take on any value, leading to infinitely many solutions. During Gaussian elimination, we identify free variables as those that do not correspond to leading ones (pivots) in their respective columns.

In the system we're working with, the variable \(z\) emerged as a free variable. This means it can assume any real number, and for each such value, there are corresponding values for \(w\), \(x\), and \(y\).

The presence of a free variable indicates that the system has an infinite number of solutions. Identifying and understanding free variables is crucial in comprehending the nature of solutions in linear algebra puzzles.