Problem 34 - Precalculus Mathematics for Calculus

Question

Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$\left\\{\begin{array}{rr} -2 x+6 y-2 z= & -12 \\ x-3 y+2 z= & 10 \\ -x+3 y+2 z= & 6 \end{array}\right.$$

Step-by-Step Solution

Answer

The system is dependent with solutions \((x, y, z) = (10 + 3y - 2t, y, t)\).

Step 1: Represent the System as a Matrix

The first step is to write down the coefficient matrix from the given system of equations. The system is:\[-2x + 6y - 2z = -12\]\[x - 3y + 2z = 10\]\[-x + 3y + 2z = 6\]The augmented matrix, which includes the coefficients and the constants, is:\[\begin{bmatrix}-2 & 6 & -2 & | & -12 \1 & -3 & 2 & | & 10 \-1 & 3 & 2 & | & 6\end{bmatrix}\]

Step 2: Use Row Operations to Simplify the Matrix

Perform row operations to simplify the matrix to row-echelon form. First, let's clean up the first column:- Swap Row 1 and Row 2 to make the first entry of the matrix a 1 (for easier calculations):\[\begin{bmatrix}1 & -3 & 2 & | & 10 \-2 & 6 & -2 & | & -12 \-1 & 3 & 2 & | & 6\end{bmatrix}\]- Add 2 times Row 1 to Row 2:\[\begin{bmatrix}1 & -3 & 2 & | & 10 \0 & 0 & 2 & | & 8 \-1 & 3 & 2 & | & 6\end{bmatrix}\]- Add Row 1 to Row 3:\[\begin{bmatrix}1 & -3 & 2 & | & 10 \0 & 0 & 2 & | & 8 \0 & 0 & 4 & | & 16\end{bmatrix}\]

Step 3: Simplify Further to Identify Consistency or Dependency

Continue simplifying the matrix:- Subtract 2 times Row 2 from Row 3 to eliminate the third column in Row 3:\[\begin{bmatrix}1 & -3 & 2 & | & 10 \0 & 0 & 2 & | & 8 \0 & 0 & 0 & | & 0\end{bmatrix}\]Notice that Row 3 now contains all zeros. This indicates that the system does not introduce a contradiction and thus is not inconsistent.

Step 4: Interpret the Simplified Matrix and Determine Relationship

The matrix indicates that we have two independent equations and a free variable (since Row 3 is all zeros). The system is thus dependent and has infinitely many solutions. The free variable can be chosen freely, and the other variables will depend on it.

Step 5: Express the Solution in Terms of the Free Variable

From the reduced matrix:- Let \( z = t \) (where \( t \) is any real number, acting as a free variable).- From the second row: \( 2z = 8 \Rightarrow z = 4 \times (t = 4 \rightarrow z = t + 4) \).- From the first row, express \( x \) in terms of \( t \): \[ x - 3y + 2z = 10 \] Substitute \( z = t \): \[ x - 3y + 2t = 10 \] Solve for \( x \): \[ x = 10 + 3y - 2t \]Hence, the solution can be expressed as \((x, y, z) = (10 + 3y - 2t, y, t)\), where \(t\) and \(y\) are free parameters.

Key Concepts

Matrix Representation

In systems of linear equations, representing the system as a matrix simplifies manipulation and helps visualize the equations' structure. Each equation in the system is translated into a row within the matrix, while the unknowns correspond to columns. The matrix representation for the system of equations given is structured as an augmented matrix:

• The coefficients of the variables form the left part of the matrix.
• The constant terms from the right side of the equations form the extra column to the right, separated by a line.

The provided exercise translates into the matrix:\[\begin{bmatrix}-2 & 6 & -2 & | & -12 \1 & -3 & 2 & | & 10 \-1 & 3 & 2 & | & 6\end{bmatrix}\]This representation allows for easier computation steps like row operations conducted next.

Row Operations

Row operations are fundamental tools in matrix algebra for solving systems of linear equations. They help transform matrices into simpler forms such as row-echelon form or reduced row-echelon form. The main types of row operations include:

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

In the example, the matrix was manipulated:- Rows were swapped and adjusted to simplify calculations.- The process continued until the matrix reached a form that provided clear insight into the existence of solutions.These operations allowed the system to be reshaped to:\[\begin{bmatrix}1 & -3 & 2 & | & 10 \0 & 0 & 2 & | & 8 \0 & 0 & 0 & | & 0\end{bmatrix}\]Reaching this form reveals dependencies or inconsistencies among the equations.

Dependent System

A system of linear equations is dependent when it has an infinite number of solutions. This occurs when one or more of the equations can be expressed as linear combinations of the others. As seen in the exercise, after employing row operations, the final row of the matrix consisted entirely of zeros:\[0x + 0y + 0z = 0\]This does not produce a new equation, indicating dependency within the system. Here, the system allowed for free variables where specific variables could take on any value, with other variables adjusted accordingly. In the solution, variables such as:

• \( z = t \)
• \( x = 10 + 3y - 2t \)

Are used to express the dependent solutions with \( t \) as a free variable.

Inconsistent System

An inconsistent system occurs when no solution satisfies all the equations simultaneously. Typical indicators include encountering contradictions such as:

• Rows like \( 0x + 0y + 0z = 1 \) appearing in the matrix's reduced form.

In the given problem, after performing all necessary row operations, all equations aligned without contradictions, resulting in one of the rows being entirely zeros. This confirmed that instead of being inconsistent, the system was dependent. Remember, inconsistent systems and dependent systems are critical concepts in linear algebra. Recognizing these patterns helps in understanding solutions' structures and assessing feasibility.