Problem 4 - Precalculus Mathematics for Calculus

Question

The augmented matrix of a system of linear equations is given in reduced row- echelon form. Find the solution of the system. $$\begin{aligned} &\text { (a) }\left[\begin{array}{llll} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 3 \end{array}\right]\\\ &x=\text{____}\\\ &y=\text{____}\\\ &z=\text{____} \end{aligned}$$ $$\begin{aligned} &\text { (b) }\left[\begin{array}{llll} 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\\\ &\boldsymbol{x}=\\\ &y=\text{____}\\\ &z=\text{____} \end{aligned}$$ $$\begin{aligned} &\text { (c) }\left[\begin{array}{llll} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 3 \end{array}\right]\\\ &x=\text{____}\\\ &y=\text{____}\\\ &z=\text{____} \end{aligned}$$

Step-by-Step Solution

Answer

(a) \(x = 2, y = 1, z = 3\). (b) \(x = 2-t, y = 1-t, z = t\). (c) No solution.

Step 1: Identify Variables for Part A

The reduced row-echelon form matrix for part (a) is given as \[\begin{bmatrix} 1 & 0 & 0 & 2 \0 & 1 & 0 & 1 \0 & 0 & 1 & 3 \end{bmatrix}\]. This matrix represents the system of equations: \(x = 2\), \(y = 1\), and \(z = 3\). The solution is directly given as the entries in the last column.

Step 2: Identify Variables for Part B

The reduced row-echelon form matrix for part (b) is given as \[\begin{bmatrix} 1 & 0 & 1 & 2 \0 & 1 & 1 & 1 \0 & 0 & 0 & 0 \end{bmatrix}\]. This represents the following system of equations: \(x + z = 2\) and \(y + z = 1\). The third row \(0 = 0\) indicates a free variable. Set \(z = t\) (a parameter) and solve for \(x\) and \(y\): - From \(x + z = 2\), we find \(x = 2 - t\).- From \(y + z = 1\), we find \(y = 1 - t\).

Step 3: Identify Variables for Part C

The reduced row-echelon form matrix for part (c) is given as \[\begin{bmatrix} 1 & 0 & 0 & 2 \0 & 1 & 0 & 1 \0 & 0 & 0 & 3 \end{bmatrix}\]. This system of equations presents \(x = 2\) and \(y = 1\) with the third row \(0 = 3\), which is a contradiction implying the system has no solution. A row like \(0 = 3\) indicates inconsistency, hence there is no solution.

Key Concepts

Reduced Row-Echelon Form

Understanding the reduced row-echelon form (RREF) is crucial for solving systems of linear equations. This matrix form is derived by using a series of elementary row operations to simplify a matrix. In the RREF, each leading entry (also called pivot) in a row is 1, and all entries above and below each pivot are zeros. Additionally, each pivot is to the right of the pivot in the row above it.

Consider the matrix for part (a):

• Row 1: The leading entry is `1` in the `x` column.
• Row 2: The leading entry is `1` in the `y` column.
• Row 3: The leading entry is `1` in the `z` column.

This setup clearly shows each variable in its own equation without any mixed variables present. Thus, it's straightforward to read the solution directly from the matrix.

The RREF provides valuable information quickly, especially when it comes to identifying directly what values variables have, if they exist. For matrices, like in part (c), when one of the rows results in an equation like `0 = 3`, it indicates an inconsistency, showing there are no solutions for that system.

System of Linear Equations

A system of linear equations is essentially a collection of multiple equations that we solve simultaneously. These systems can be represented in matrix form, like the augmented matrix you've dealt with above. This representation aids in applying algebraic methods to find solutions.

For instance, in part (a) and part (b), the rows of the matrix represent linear equations in terms of `x`, `y`, and `z`.

• Part (a) yields three distinct equations; each equation gives a specific solution for each variable directly.
• Part (b) has a third row of all zeros, indicating that there's a free variable involved. This reflects that the system has infinitely many solutions, parametrized by setting one variable free, often denoted as `t`.
• Part (c), on the other hand, reflects an inconsistent system with `0 = 3`, meaning these equations cannot coexist with any set of variables satisfying them simultaneously.

The key takeaway with systems of linear equations is whether they are consistent (having at least one solution), inconsistent (no solution), or dependent (infinitely many solutions), which can be directly determined from their RREF.

Solution of the System

Finding the solution of a system presented in RREF involves interpreting the matrix back into equations and solving them. Let's explore how solutions are found using the matrices given in the exercise.

For part (a): We have a unique solution. The matrix trivially converts to equations: `x = 2`, `y = 1`, `z = 3`.

In part (b): The presence of a zero row leads to a discussion on free variables. The solution is:

• Express `z` as a parameter `t`.
• Solve the equations `x + z = 2` as `x = 2 - t` and `y + z = 1` as `y = 1 - t`.

This shows that there are infinitely many solutions depending on `t`.

Part (c) presents an inconsistency (`0 = 3` row), hence no solution exists. An inconsistent system means that no values of `x`, `y`, and `z` satisfy all equations simultaneously. Recognizing these distinctions between unique, infinitely many, or no solutions helps immensely in understanding the nature of the systems of equations you're dealing with.