Problem 21 - Differential Equations and Linear Algebra

Question

Determine the solution set to the sys\(\operatorname{tem} A \mathbf{x}=\mathbf{b}\) for the given coefficient matrix \(A\) and right-hand side vector b. $$A=\left[\begin{array}{rrr} 0 & 1 & -1 \\ 0 & 5 & 1 \\ 0 & 2 & 1 \end{array}\right], \mathbf{b}=\left[\begin{array}{r} -2 \\ 8 \\ 5 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution set for the given system $A\mathbf{x}=\mathbf{b}$ is {(0, 1, 3)}, obtained by performing Gaussian elimination on the augmented matrix and solving the resulting equations.

Step 1: Analyze the coefficient matrix A and vector b

Here, the given coefficient matrix A and vector b are: \[A=\left[\begin{array}{rrr} 0 & 1 & -1 \\ 0 & 5 & 1 \\ 0 & 2 & 1 \end{array}\right], \mathbf{b}=\left[\begin{array}{r} -2 \\ 8 \\ 5 \end{array}\right]\] First, we check the consistency of the system. Since matrix A does not have a row of all zeroes, the system is consistent.

Step 2: Perform Gaussian elimination to find the row echelon form

To solve this system, we need to perform Gaussian elimination on the augmented matrix \([A|\mathbf{b}]\): \[\left[\begin{array}{rrr|r} 0 & 1 & -1 & -2 \\ 0 & 5 & 1 & 8 \\ 0 & 2 & 1 & 5 \end{array}\right]\] To start with, we will eliminate the elements below the pivot in the second column: Divide Row 2 by 5 and then subtract Row 1 multiplied by 2 from Row 3. \[\left[\begin{array}{rrr|r} 0 & 1 & -1 & -2 \\ 0 & 1 & 0.2 & 1.6 \\ 0 & 0 & 3 & 9 \end{array}\right]\] Now we have obtained the row echelon form.

Step 3: Perform back-substitution to find the solution

The row echelon form of the augmented matrix represents the following system of equations: \[ \begin{cases} y - z = -2 \\ y + 0.2z = 1.6 \\ 3z = 9 \end{cases} \] Solve the third equation for z: \[z = \frac{9}{3} = 3\] Now substitute the value of z into the second equation to find the value of y: \[y + 0.2(3) = 1.6\] \[y = 1.6 - 0.6 = 1\] Now substitute the values of y and z into the first equation to find the value of x: \[1 - 3 = -2\] The unique solution for the system is: \[\mathbf{x} = \left[\begin{array}{r} 0 \\ 1 \\ 3 \end{array}\right]\] Thus, the solution set for the given system is {(0, 1, 3)}.

Key Concepts

Gaussian Elimination

Gaussian elimination is a method used for solving systems of linear equations. It transforms a matrix into a simpler form, similar to a step-by-step simplification. The process starts with an augmented matrix, which combines the coefficients and the constants from the equations. The main goal here is to turn the system into a form that is easier to solve.

• Start by identifying a pivot, which is a non-zero element in the matrix that you can use to eliminate other values in its column.
• Apply row operations to replace these elements below the pivot with zeros.

In our original example, only row operations on non-zero elements were needed because the first column consisted solely of zeros. Working with columns can sometimes be a bit tricky, so it's crucial to make sure each step is correctly carried out.
At the end of Gaussian elimination, you are left with a transformed matrix that lays the foundation for the subsequent steps.

Row Echelon Form

After performing Gaussian elimination, a matrix is in row echelon form. This format simplifies solving equations because of its structured appearance. A matrix achieves this form when:

• All non-zero rows are above any rows of all zeros.
• The leading entry (or pivot) in each non-zero row is 1.
• The leading entry in a row is to the right of the leading entry in the previous row.

In the example provided, the transformation process led to the simplified system:
1. \(y - z = -2\)
2. \(y + 0.2z = 1.6\)
3. \(3z = 9\)
These simplifications make it easier to identify which variables can be solved directly, speeding up the solving process.

Back-Substitution

Once the matrix is in row echelon form, back-substitution is the method used to find the specific solution by solving the equations from bottom to top. You start from the lowest variable and move upwards.

• Solve the equation for the known variable; in our example, solve for \(z\) from \(3z = 9\): \(z = 3\).
• Substitute the found variable into the earlier equation: \(y + 0.2z = 1.6\).
• Clarify this equation to find \(y\): \(y = 1\).
• Finally, use \(y\) and \(z\) in the first equation to find \(x\).

The work is sequential and relied on the decisions made during Gaussian elimination's steps and row echelon form conversion. This approach is systematic, making it easier to check your work and ensure precision.

Consistency of Systems

Consistency in a system of linear equations means there exists at least one set of values for the variables that satisfy all the equations. Analyzing the original matrix provides a crucial step in understanding the nature of the solutions.

• If a row consists only of 0s in the coefficient matrix, but there's a non-zero number on the right-hand side, it generally indicates inconsistency, meaning no solutions exist.
• Our example showed no such row, suggesting that the system is consistent—as noted in Step 1.

Consistency is an essential check even before proceeding with Gaussian elimination, ensuring the effort of transformation and back-substitution will lead to an actual solution. Modern adjustments or simplifications in a matrix's form initially guide through analyzing for potential trails of solutions.