Problem 30 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

The system has infinitely many solutions, given by \((x, y, z) = (2 - 2y, y, 1)\), where \(y\) is a free parameter.

Step 1: Write the system as a matrix equation

Start by writing the system of equations in matrix form: \[\begin{bmatrix}2 & 4 & -1 \1 & 2 & 4 \1 & 2 & -2\end{bmatrix}\begin{bmatrix} x \ y \ z \end{bmatrix} =\begin{bmatrix} 3 \ 6 \ 0 \end{bmatrix}\]

Step 2: Form the augmented matrix

Combine the coefficient matrix with the constants to form the augmented matrix:\[\begin{bmatrix}2 & 4 & -1 & | & 3 \1 & 2 & 4 & | & 6 \1 & 2 & -2 & | & 0\end{bmatrix}\]

Step 3: Apply row operations to simplify

Use row operations to simplify the augmented matrix. First, subtract the first row from the second and third rows:\[R_2 \to R_2 - 0.5R_1 \R_3 \to R_3 - 0.5R_1\]This results in the matrix:\[\begin{bmatrix}2 & 4 & -1 & | & 3 \0 & 0 & 4.5 & | & 4.5 \0 & 0 & -1.5 & | & -1.5\end{bmatrix}\]

Step 4: Further simplify the matrix

Now, add \(\frac{1}{3}\) of row 3 to row 2 to simplify:\[R_2 \to R_2 + 3R_3\]This modifies the matrix to:\[\begin{bmatrix}2 & 4 & -1 & | & 3 \0 & 0 & 0 & | & 0 \0 & 0 & -1.5 & | & -1.5\end{bmatrix}\]

Step 5: Solve for the variables

Since the only non-zero row remaining in the reduced system is \([0, 0, -1.5 | -1.5]\), we have the equation \(-1.5z = -1.5\), which simplifies to \(z = 1\). Now, using the first equation \(2x + 4y - z = 3\) with \(z = 1\), we plug in and solve for \(x\) and \(y\):\[2x + 4y - 1 = 3 \2x + 4y = 4\]. This equation can be further simplified by plugging values for \(x\) and expressing \(y\) as a parameter hold.

Step 6: Express the complete solution

The solution can be expressed in terms of any dependent variable. Choose \(x = k - 2y\) for the general solution. So one possible solution is expressed as:\[(x, y, z) = (2 - 2y, y, 1)\]where \(y\) is a free parameter.

Key Concepts

Augmented Matrix

An augmented matrix is a crucial tool when solving systems of linear equations. It combines the coefficient matrix with the constants on the right-hand side of the equations. This forms a single matrix which simplifies operations and calculations.
To create an augmented matrix, write down the coefficients of each variable from your equations into rows, aligning variables into columns.
Next to this coefficient matrix, add a vertical line followed by the constants from the equations.
For example, if you have the system:

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

The augmented matrix representation is:\[ \begin{bmatrix} 2 & 4 & -1 & | & 3 \ 1 & 2 & 4 & | & 6 \ 1 & 2 & -2 & | & 0 \end{bmatrix} \] This matrix helps you visualize the equations as a single entity, making subsequent manipulations using row operations more intuitive.

Row Operations

Row operations are techniques used to manipulate an augmented matrix, helping to simplify it and find solutions to linear equations. There are three main row operations:

• Swapping two rows. This helps rearrange the matrix without changing the solution.
• Multiplying a row by a nonzero scalar. This can simplify calculations by adjusting coefficients.
• Adding or subtracting a multiple of one row to another. This operation is essential for eliminating variables in certain rows.

For example, after forming the augmented matrix with the given system, the initial steps involve subtracting a scaled version of one row from others to zero out coefficients of certain variables. This leads to:\[ R_2 \rightarrow R_2 - 0.5R_1 \] \[ R_3 \rightarrow R_3 - 0.5R_1 \] These row operations gradually reduce the system to a simpler form that can be solved easily, making it possible to isolate variables.

Parametric Solution

A parametric solution provides a way to express the solutions of a system of equations where not all variables are independent. This often happens when you have fewer equations than variables, leading to infinite solutions.
In these cases, some variables are expressed in terms of one or more freely chosen parameters.
In our example, after reducing the matrix, we find that:- The variable \(z\) is independent with a specific value (\(z = 1\))- The other variables, \(x\) and \(y\), are dependent on each other
The equation \(2x + 4y = 4\) suggests that by choosing any value for \(y\), you can derive a corresponding \(x\). This leads to the parametric form:\((x, y, z) = (2 - 2y, y, 1)\)Here, \(y\) acts as the free parameter, allowing the solution set to be described as an infinite line in three-dimensional space. Each unique \(y\) value generates a specific solution for \(x\), thereby forming a family of solutions rather than just one.