Problem 32 - Precalculus Mathematics for Calculus

Question

Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$\left\\{\begin{array}{rr} x-2 y+5 z= & 3 \\ -2 x+6 y-11 z= & 1 \\ 3 x-16 y-20 z= & -26 \end{array}\right.$$

Step-by-Step Solution

Answer

The system is dependent, with solution \( x = 10, \ y = \frac{7}{2}, \ z = 0 \).

Step 1: Write the system in matrix form

First, express the system of linear equations as an augmented matrix. The system is: \[ \begin{bmatrix} 1 & -2 & 5 & | & 3 \ -2 & 6 & -11 & | & 1 \ 3 & -16 & -20 & | & -26 \end{bmatrix} \]

Step 2: Perform row operations to obtain Row Echelon Form (REF)

Use row operations to simplify the matrix:1. Use row 1 to eliminate the first element of row 2: \[ R_2 = R_2 + 2R_1 \rightarrow (0, 2, -1, |, 7) \]2. Use row 1 to eliminate the first element of row 3: \[ R_3 = R_3 - 3R_1 \rightarrow (0, -10, -35, |, -35) \]Now the matrix is:\[ \begin{bmatrix} 1 & -2 & 5 & | & 3 \ 0 & 2 & -1 & | & 7 \ 0 & -10 & -35 & | & -35 \end{bmatrix} \]

Step 3: Simplify further to obtain Reduced Row Echelon Form (RREF)

3. Use row 2 to eliminate the second element of row 3: Solve row 2 as a pivot row: \[ R_2 = \frac{1}{2}R_2 \rightarrow (0, 1, -\frac{1}{2}, |, \frac{7}{2}) \] 4. Substitute it into row 3: \[ R_3 = R_3 + 10 \times R_2 \rightarrow (0, 0, -40, |, 0) \] Let's update the matrix:\[ \begin{bmatrix} 1 & -2 & 5 & | & 3 \ 0 & 1 & -\frac{1}{2} & | & \frac{7}{2} \ 0 & 0 & -40 & | & 0 \end{bmatrix} \]

Step 4: Interpret the RREF result

Notice that row 3 simplifies to: \[ 0x + 0y - 40z = 0 \rightarrow z = 0 \] Now back-substitute to find the solutions for other variables in terms of free variables. Since row 3 yields a non-trivial solution and row 1, row 2 are dependent, the original system is dependent.

Step 5: Identify the free variable and express complete solution

From row 2: \[ y - \frac{1}{2}z = \frac{7}{2} \rightarrow y = \frac{7}{2} + \frac{1}{2}z \]Substitute the known value of \( z = 0 \):\[ y = \frac{7}{2} \]From row 1:\[ x - 2y + 5z = 3 \rightarrow x - 2 \times \frac{7}{2} = 3 \]\[ x - 7 = 3 \rightarrow x = 10 \] Thus the solution is: \[ x = 10, \quad y = \frac{7}{2}, \quad z = 0 \]

Step 6: Conclusion: Dependent Solution

The given system is dependent with the solution \( x = 10, \ y = \frac{7}{2}, \ z = 0 \).

Key Concepts

Row Echelon Form

The Row Echelon Form (REF) is a way to structure a matrix to aid in solving systems of linear equations. To put a matrix in REF, we use row operations until it satisfies a few key properties:

• All zero rows (if any) are at the bottom of the matrix.
• Every leading non-zero element, called a pivot, is to the right of the pivots in the rows above.
• All elements below each pivot are zeros.

To achieve REF from our system of equations, we start with the augmented matrix form, as shown in step 1 of the original solution. We then perform operations like adding or subtracting multiples of rows to zero out entries below the pivots. For example, we eliminate the first element of rows 2 and 3 by adjusting them with respect to row 1. As a result, the matrix progresses towards a triangular shape. This form is helpful as it prepares the matrix for further simplification into Reduced Row Echelon Form (RREF) or can directly be used to back-substitute to find the solution.

Reduced Row Echelon Form

Moving from Row Echelon Form (REF) to Reduced Row Echelon Form (RREF) involves additional simplifications that make solving the system of equations straightforward. In RREF, the matrix meets the criteria of REF and adds more conditions:

• Pivot positions are the only non-zero entries in their column.
• Pivots are set to 1 for easier back-substitution.

The transition to RREF often involves additional row operations to clear non-zero entries above the pivots. In our exercise, once REF is achieved, we continue to manipulate the matrix so that each leading coefficient is 1 and the pivots become the sole non-zero elements in their columns. This is done by normalizing rows to make the pivot 1 and then eliminating other entries in those pivot columns. A correctly reduced row echelon matrix makes it clear which variables are determined by the system and which might be free variables, signaling potential system dependency.

Dependent System

In the context of linear algebra, a system of equations is called dependent when there are infinitely many solutions. Dependency usually arises when at least one of the equations is a linear combination of the others, meaning there isn’t enough independent information to pinpoint a unique solution. In our particular problem, RREF demonstrated this when row 3 simplified to zero, revealing the relationship among variables necessary for dependency. As a result, certain variables are not bound to a single value and are instead free to vary. In this exercise, after solving for the dependent system, we found a specific solution but acknowledged potential variability of another solution meeting the dependency condition. Understanding dependencies is critical, as it highlights the importance of each equation in uniquely determining a solution set.