Problem 18 - Differential Equations and Linear Algebra

Question

Use Gauss-Jordan elimination to determine the solution set to the given system. $$\begin{aligned} 2 x_{1}-x_{2}+3 x_{3}+x_{4}-x_{5} &=11 \\ x_{1}-3 x_{2}-2 x_{3}-x_{4}-2 x_{5} &=2 \\ 3 x_{1}+x_{2}-2 x_{3}-x_{4}+x_{5} &=-2 \\ x_{1}+2 x_{2}+x_{3}+2 x_{4}+3 x_{5} &=-3 \\ 5 x_{1}-3 x_{2}-3 x_{3}+x_{4}+2 x_{5} &=2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution set for the given system of linear equations is: $$\begin{aligned} x_{1} &= -\frac{1}{3}\left(0.5t -7.5\right) +\frac{5}{3}t + 7 \\ x_{2} &= 1.5\left(-\frac{1}{3}\left(0.5t -7.5\right) +\frac{4}{3}t +\frac{43}{7}\right) - 0.5\left(0.5t - 7.5\right) -\frac{1}{2}t -5.5 \\ x_{3} &= -\frac{1}{3}\left(0.5t -7.5\right) + \frac{4}{3}t +\frac{43}{7} \\ x_{4} &= 0.5t - 7.5 \\ x_{5} &= t \end{aligned}$$ where \(t\) is a free variable that can take any real value.

Step 1: Write the augmented matrix

First, we will write the given system of linear equations as an augmented matrix. \( \left[ \begin{array}{ccccc|c} 2 & -1 & 3 & 1 & -1 & 11 \\ 1 & -3 & -2 & -1 & -2 & 2 \\ 3 & 1 & -2 & -1 & 1 & -2 \\ 1 & 2 & 1 & 2 & 3 & -3 \\ 5 & -3 & -3 & 1 & 2 & 2 \end{array} \right] \)

Step 2: Apply row operations to simplify the matrix

We will now apply a series of row operations to simplify the augmented matrix and obtain an equivalent system of linear equations. 1. Swap row 1 and row 2 2. Subtract 1 times row 1 from row 3 3. Subtract 1 times row 1 from row 4 4. Subtract 5 times row 1 from row 5 \( \left[ \begin{array}{ccccc|c} 1 & -3 & -2 & -1 & -2 & 2 \\ 2 & -1 & 3 & 1 & -1 & 11 \\ 0 & 7 & 0 & 0 & 5 & -8 \\ 0 & 5 & 3 & 3 & 5 & -5 \\ 0 & 12 & 7 & 6 & 12 & -8 \end{array} \right] \) 5. Replace row 2 with (1/2)*row 2 6. Add 3 times row 2 to row 1 7. Subtract 7 times row 2 from row 3 8. Subtract 5 times row 2 from row 4 9. Subtract 12 times row 2 from row 5 \( \left[ \begin{array}{ccccc|c} 1 & 0 & 1 & 0 & -2 & 7 \\ 0 & 1 & -1.5 & -0.5 & 0.5 & -5.5\\ 0 & 0 & 10.5 & 3.5 & 8.5 & 30.5 \\ 0 & 0 & 10.5 & 5.5 & 7.5 & 22.5 \\ 0 & 0 & 25 & 12 & 18 & 58 \end{array} \right] \) 10. Replace row 3 with (2/21)*row 3 11. Subtract row 3 from row 4 12. Subtract 5 times row 3 from row 5 \( \left[ \begin{array}{ccccc|c} 1 & 0 & 1 & 0 & -2 & 7 \\ 0 & 1 & -1.5 & -0.5 & 0.5 & -5.5\\ 0 & 0 & 1 & 1/3 & 4/3 & 43/7 \\ 0 & 0 & 0 & 2 & -1 & -15 \\ 0 & 0 & 0 & 2.5 & -2.5 & -15 \end{array} \right] \) 13. Replace row 4 with (1/2)*row 4 14. Subtract 2.5 times row 4 from row 5 \( \left[ \begin{array}{ccccc|c} 1 & 0 & 1 & 0 & -2 & 7 \\ 0 & 1 & -1.5 & -0.5 & 0.5 & -5.5\\ 0 & 0 & 1 & 1/3 & 4/3 & 43/7 \\ 0 & 0 & 0 & 1 & -0.5 & -7.5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \)

Step 3: Analyse the simplified matrix

Now, we can observe the matrix and write down the equivalent system of linear equations: $$\begin{aligned} x_{1} + x_{3} - 2 x_{5} &= 7 \\ x_{2} - 1.5 x_{3} - 0.5 x_{4} + 0.5 x_{5} &= -5.5 \\ x_{3} + \frac{1}{3} x_{4} + \frac{4}{3} x_{5} &= \frac{43}{7} \\ x_{4} - 0.5 x_{5} &= -7.5 \\ 0 &= 0 \end{aligned}$$ To express the solution in terms of free variables, let \(x_{5} = t\), where \(t\) is a free variable. Then, we can solve for the other variables as follows: $$\begin{aligned} x_{4} &= 0.5t - 7.5 \\ x_{3} &= -\frac{1}{3}\left(0.5t -7.5\right) +\frac{4}{3}t +\frac{43}{7} \\ x_{2} &= 1.5\left(-\frac{1}{3}\left(0.5t -7.5\right) +\frac{4}{3}t +\frac{43}{7}\right) + 0.5\left(0.5t - 7.5\right) -0.5t -5.5 \\ x_{1} &= -1\left(-\frac{1}{3}\left(0.5t -7.5\right) +\frac{4}{3}t +\frac{43}{7}\right) + 2t + 7 \end{aligned}$$ Thus, the solution set is given by: $$\begin{aligned} x_{1} &= -\frac{1}{3}\left(0.5t -7.5\right) +\frac{5}{3}t + 7 \\ x_{2} &= 1.5\left(-\frac{1}{3}\left(0.5t -7.5\right) +\frac{4}{3}t +\frac{43}{7}\right) - 0.5\left(0.5t - 7.5\right) -\frac{1}{2}t -5.5 \\ x_{3} &= -\frac{1}{3}\left(0.5t -7.5\right) + \frac{4}{3}t +\frac{43}{7} \\ x_{4} &= 0.5t - 7.5 \\ x_{5} &= t \end{aligned}$$ where \(t\) is a free variable that can take any real value.

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool in linear algebra. It represents a system of linear equations in a compact form. By using an augmented matrix, we combine the coefficients of the variables and the constants from the equations into one matrix.
In the system of linear equations, each row of the augmented matrix corresponds to one equation. For instance, in the given system, the set of equations is transformed into an augmented matrix as follows:

• The matrix's first five columns contain the coefficients of the variables \(x_1\), \(x_2\), \(x_3\), \(x_4\), and \(x_5\).
• The column after the vertical bar holds the constants from each equation.

This simplified representation allows us to use systematic operations to solve for the variables. Augmented matrices are integral in solving systems using methods like Gauss-Jordan elimination.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. The solution to this system is the set of values for the variables that makes all the equations true simultaneously.
For the given set of equations:

• We have five unknowns: \(x_1, x_2, x_3, x_4, \) and \(x_5\).
• Each linear equation provides a linear relationship among these unknowns.

The objective is to find values for these variables that satisfy all the equations at once. This involves conversion into an augmented matrix and using methods like Gauss-Jordan elimination to systematically reduce and find solutions. Without transformation into a matrix form, solving such a complex system manually might be error-prone and time-consuming.

Row Operations

Row operations are the backbone of matrix manipulation during Gauss-Jordan elimination. They help to systematically transform the matrix into a simpler form called the row-reduced echelon form (RREF).
There are three main types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row from another row.

These operations maintain the equivalence of the original system while simplifying the matrix iteratively. For instance, during the solution process, rows were swapped, scaled, and combined to isolate variables and ultimately simplify the augmented matrix. By the end of these transformations, we obtain a diagonal form where each non-zero row corresponds to a leading variable in terms of others or even directly gives the solution.

Free Variable

In a system of linear equations, a free variable is one that can take any real value, showing flexibility in the solution set. This occurs when there are more variables than independent equations.
Consider the provided solution steps:

• Variable \(x_5\) was chosen as the free variable.
• It doesn't have a specific value fixed by the independent equations.
• The other variables \(x_1, x_2, x_3,\) and \(x_4\) are expressed in terms of this free variable \(x_5\).

Free variables provide systems with infinite solutions. In practical terms, for each real value \(t\) taken by \(x_5\), we find a specific combination of values for the other variables, representing a line or plane in multidimensional space depending on the system's dimensions.