Problem 35 - Precalculus: Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. \(\left\\{\begin{aligned} x+3 y-2 z &=0 \\ 2 x+\quad & 4 z=4 \\ 4 x+6 y &=4 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The system has infinitely many solutions with parameters: \((x, y, z) = (-2 - 2z, \frac{2 + 4z}{3}, z)\).

Step 1: Write the System of Equations in a Matrix Form

First, we express the given linear system in matrix form. The system consists of three equations: \[ \begin{align*} x + 3y - 2z &= 0, \ 2x + 4z &= 4, \ 4x + 6y &= 4. \end{align*} \] This can be rewritten in augmented matrix form as: \[ \left[ \begin{array}{ccc|c} 1 & 3 & -2 & 0 \ 2 & 0 & 4 & 4 \ 4 & 6 & 0 & 4 \end{array} \right]. \]

Step 2: Use Row Operations to Simplify the Matrix

Our goal is to transform the augmented matrix into row-echelon form using row operations. Start by focusing on the first column to eliminate elements below the first pivot (1):1. Subtract 2 times the first row from the second row (R2): \[ R2 \leftarrow R2 - 2R1 \rightarrow \left[ \begin{array}{ccc|c} 1 & 3 & -2 & 0 \ 0 & -6 & 8 & 4 \ 4 & 6 & 0 & 4 \end{array} \right]. \]2. Subtract 4 times the first row from the third row (R3): \[ R3 \leftarrow R3 - 4R1 \rightarrow \left[ \begin{array}{ccc|c} 1 & 3 & -2 & 0 \ 0 & -6 & 8 & 4 \ 0 & -6 & 8 & 4 \end{array} \right]. \]

Step 3: Find a Leading 1 in the Second Column If Possible

Observe that rows 2 and 3 are identical. To simplify our matrix further:1. Subtract row 2 from row 3: \[ R3 \leftarrow R3 - R2 \rightarrow \left[ \begin{array}{ccc|c} 1 & 3 & -2 & 0 \ 0 & -6 & 8 & 4 \ 0 & 0 & 0 & 0 \end{array} \right]. \]The third row becoming all zeros indicates a parameter is involved or the system may have infinite solutions.

Step 4: Solve the Simplified System

Convert the resulting row-echelon matrix back into equations:1. From row 1: \[ x + 3y - 2z = 0. \]2. From row 2: \[ -6y + 8z = 4 \quad \implies y = \frac{4 + 8z}{6} = \frac{2 + 4z}{3}. \]By substituting the expression for \(y\) into the equation from row 1, solve for \(x\): \[ x + 3\left(\frac{2 + 4z}{3}\right) - 2z = 0 \quad \implies x + 2 + 4z - 2z = 0 \quad \implies x = -2 - 2z. \]

Step 5: Express the Solution in Parametric Form

The solution to the system is based on the parameter \(z\) representing free variables. We express \(x\), \(y\), and \(z\) in terms of \(z\):- \( x = -2 - 2z \),- \( y = \frac{2 + 4z}{3} \),- \( z = z \).Thus, the solution is:\[ (x, y, z) = (-2 - 2z, \frac{2 + 4z}{3}, z), \] where \(z\) is any real number.

Key Concepts

Matrix Form

To solve a system of linear equations, we can use the matrix form, which simplifies the process significantly. The given system of equations includes:

• \(x + 3y - 2z = 0\)
• \(2x + 4z = 4\)
• \(4x + 6y = 4\)

By writing these equations in matrix form, we translate the system into a structured array of numbers, known as an augmented matrix. This transformation embodies the coefficients of the variables and the constants on the right-hand side as follows:\[\begin{array}{ccc|c}1 & 3 & -2 & 0 \2 & 0 & 4 & 4 \4 & 6 & 0 & 4\end{array}\]The first three columns comprise the coefficients of \(x\), \(y\), and \(z\), while the fourth is the corresponding constant from each equation. Using augmented matrices allows us to harness efficient techniques that simplify the system step by step.

Row Operations

Once the system is expressed in matrix form, we use row operations to simplify it, aiming for a form that reveals the solution more readily. Row operations consist of three main actions:

• swap two rows,
• multiply a row by a non-zero scalar,
• add or subtract the multiple of one row from another.

In our solution, we focused initially on eliminating numbers below the first element or pivot in the matrix's first column:

• Subtracting twice the first row from the second row simplifies this to: \[\begin{array}{ccc|c}1 & 3 & -2 & 0 \0 & -6 & 8 & 4 \4 & 6 & 0 & 4 \end{array}\]
• Subtracting four times the first row from the third row makes it: \[\begin{array}{ccc|c}1 & 3 & -2 & 0 \0 & -6 & 8 & 4 \0 & -6 & 8 & 4 \end{array}\]

Notice the duplication in rows 2 and 3. A subsequent subtraction of one from the other gives rise to a row of zeros, indicating redundancy or infinite solutions. Thus, row operations illuminate the inherent characteristics of the system.

Parametric Solution

Parametric solutions become necessary when a linear system might not have a unique solution, often due to free variables. In this system, the elimination process reveals that there are fewer independent equations than variables. This is indicated by the third row, which simplifies entirely to zeros, signifying that one variable can take many values.We find parameter-based expressions for each variable. First, rewrite the simplified row-echelon form back into equations:

• \(x + 3y - 2z = 0\)
• \(-6y + 8z = 4\)

Now isolate \(y\) in terms of \(z\): \[y = \frac{2 + 4z}{3}\]Plug this into the equation for \(x\): \[x = -2 - 2z\]Thus, our solution using the parameter \(z\) becomes a set of expressions:

• \(x = -2 - 2z\)
• \(y = \frac{2 + 4z}{3}\)
• \(z = z\)

This solution indicates that \(z\), the free variable, can assume any real number, resulting in infinitely many potential solutions.