Problem 24 - Differential Equations and Linear Algebra

Question

Determine the solution set to the given linear system of equations. $$\begin{aligned} 3 x_{1} -x_{3}+2 x_{4}-x_{5}=1, \\ x_{1}+3 x_{2}+x_{3}-3 x_{4}+2 x_{5} =-1, \\ 4 x_{1}-2 x_{2}-3 x_{3}+6 x_{4}-x_{5} =5. \\ x_{4}+4 x_{5} =-2. \end{aligned}$$

Step-by-Step Solution

Answer

The solution set to the given linear system of equations is \[ \left\{ \left(-\frac{13}{3}t + 1, \frac{5}{3}t + \frac{2}{3}, 5t - 1, -4t - 2, t\right) \mid t \in \mathbb{R} \right\}.

Step 1: Write the system in matrix form

Write the given system of linear equations as an augmented matrix: \[ \left[ \begin{array}{ccccc|c} 3 & 0 & -1 & 2 & -1 & 1 \\ 1 & 3 & 1 & -3 & 2 & -1 \\ 4 & -2 & -3 & 6 & -1 & 5 \\ 0 & 0 & 0 & 1 & 4& -2 \end{array} \right] \]

Step 2: Reduce the matrix to row echelon form

Apply Gaussian elimination to obtain the row echelon form: \[ \left[ \begin{array}{ccccc|c} 1 & 3 & 1 & 0 & -3 & -1 \\ 0 & 1 & \frac{2}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & 0 & -5 & -1 \\ 0 & 0 & 0 & 1 & 4 & -2 \end{array} \right] \]

Step 3: Apply back substitution to obtain the reduced row echelon form

Perform back substitution and obtain the reduced row echelon form: \[ \left[ \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & \frac{13}{3} & 1 \\ 0 & 1 & 0 & 0 & -\frac{5}{3} & \frac{2}{3} \\ 0 & 0 & 1 & 0 & -5 & -1 \\ 0 & 0 & 0 & 1 & 4 & -2 \end{array} \right] \]

Step 4: Identify free variables and express the solution set

From the reduced row echelon form, we can see that \(x_1\), \(x_2\), \(x_3\), and \(x_4\) correspond to pivot columns and \(x_5\) is a free variable. Let \(x_5 = t\). Then the solution set can be expressed as: \[ \begin{aligned} x_1 &= -\frac{13}{3}t + 1 \\ x_2 &= \frac{5}{3}t + \frac{2}{3} \\ x_3 &= 5t - 1 \\ x_4 &= -4t - 2 \\ x_5 &= t \end{aligned} \] Thus, the solution set to the given linear system of equations is \[ \left\{ \left(-\frac{13}{3}t + 1, \frac{5}{3}t + \frac{2}{3}, 5t - 1, -4t - 2, t\right) \mid t \in \mathbb{R} \right\}.

Key Concepts

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations. It transforms the system into a simpler form that is easier to solve.

This is done by using a series of row operations to create zeroes below the leading coefficients, also known as pivots, in the matrix. The three key row operations include:

• Swapping two rows
• Multiplying a row by a nonzero scalar
• Adding or subtracting a multiple of one row to another row

These operations help simplify the system step by step, reducing it to a point where back substitution can be applied to find solutions. Gaussian elimination is particularly efficient with the use of matrices.

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations. It combines the coefficients of the variables and the constants from the equations into one matrix. This can be written as [A|B], where A is the coefficient matrix and B is the column matrix representing the solution of the equations.

Using an augmented matrix simplifies the process of performing operations on the system, as it allows for a straightforward, visual way to apply row operations. It also makes it easier to track changes as the system moves towards its row echelon form or further into reduced row echelon form. This streamlined format is instrumental for solving systems using Gaussian elimination.

Row Echelon Form

Row Echelon Form (REF) is a form of a matrix that allows for straightforward back substitution to find solutions to a system of equations. In REF, the matrix contains zeroes below each pivot. The pivots themselves are the leading entries in their respective rows.

The process of getting a matrix to row echelon form involves performing a series of row operations to create zeros in all positions below each pivot. Once in this form, the matrix can be further transformed into Reduced Row Echelon Form (RREF), where each pivot is 1, and all values in the pivot’s column are zero, except for the pivot itself.

Row Echelon Form is central in many algebraic methods, simplifying the solution-finding process significantly.

Free Variables

Free variables arise when a system of linear equations has infinitely many solutions. In the context of matrices, these occur when a column does not contain a pivot.

In such cases, these variables can take any real value, which means they can be parameterized in terms of one or more variables. Free variables are key for expressing the general solution to a system where parameters can vary. They allow the solution to be written in a form using arbitrary constants, representing a family of solutions.

This concept is crucial for understanding the structure of solutions in linear algebra, especially in underdetermined systems where the number of variables exceeds the number of independent equations.