Question

Decide which pairs of the variables \(x, y, z, u,\) and \(v\) are determined as functions of the others by the system $$ \begin{array}{r} x+2 y+3 z+u+6 v=0 \\ 2 x+4 y+z+2 u+2 v=0 \end{array} $$ and solve for them.

Short answer

In the given system of equations, the pairs of variables that are determined as functions of the others are: 1. $v$ as a function of $z$: $v= -\frac{1}{2}z$ 2. $z$ as a function of $v$: $z= -2v$ 3. $y$ as a function of $x, z, u$, and $v$: $y= \frac{1}{2}(6z + 6v - x - u)$ 4. $u$ as a function of $x, y, z$, and $v$: $u= 6z + 6v - x - 2y$

Step-by-step solution

Identify the dependent and independent variables

In this system of equations, we can observe that variables \(x, y, z, u,\) and \(v\) are mixed within the equations. To find out which variables are dependent on others, we can first rewrite this system in matrix form as follows: $$ \begin{bmatrix} 1 & 2 & 3 & 1 & 6 \\ 2 & 4 & 1 & 2 & 2 \end{bmatrix} $$ By standard row operations, we will eliminate one of the variables, say \(x\).

Perform row operations to eliminate \(x\)

We will eliminate the variable \(x\) by multiplying the first row by 2, and then subtracting it from the second row. $$ \begin{bmatrix} 1 & 2 & 3 & 1 & 6 \\ 0 & 0 & -5 & 0 & -10 \end{bmatrix} $$ Now, we have a simplified system: $$ \begin{array}{r} x + 2y + 3z + u + 6v = 0 \\ -5z - 10v = 0 \end{array} $$

Solve for \(z\) and \(v\)

We can now see that variables \(x, y,\) and \(u\) have been eliminated from the second equation: $$-5z - 10v = 0$$ Divide by \(-5\) to get: $$z + 2v = 0$$ Thus, \(v\) is determined as a function of \(z\): $$v = -\frac{1}{2}z$$ Similarly, \(z\) is determined as a function of \(v\): $$z = -2v$$

Solve for the other variables

Now, substitute the value of \(z\) and \(v\) back into the first equation: $$x + 2y + 3(-2v) + u + 6(-\frac{1}{2}z) = 0$$ Simplify the equation: $$x + 2y - 6v + u - 6z = 0$$ Now, we can express the other pairs of variables (\(x, y\), and \(u\)) as functions of \(z\) and \(v\): $$x = 6z + 6v - 2y - u$$ From this equation, we can see now that \(y\) is determined as a function of \(x, z, u,\) and \(v\): $$y = \frac{1}{2}(6z + 6v - x - u)$$ And similarly, \(u\) is determined as a function of \(x, y, z,\) and \(v\): $$u = 6z + 6v - x - 2y$$

Conclusion

In conclusion, the pairs of variables \(x, y, z, u,\) and \(v\) that are determined as functions of the others in the given system of equations are: $$ \begin{cases} v = -\frac{1}{2}z \\ z = -2v \\ y = \frac{1}{2}(6z + 6v - x - u) \\ u = 6z + 6v - x - 2y \end{cases} $$

Key concepts

Row Operations

Row operations are fundamental tools used to manipulate a matrix in order to solve systems of linear equations. These operations consist of three types:

• Swapping two rows,
• Scaling a row by a nonzero constant,
• Adding or subtracting multiples of one row to another row.

For example, in the exercise provided, one of the row operations performed was to eliminate the variable x by multiplying the first row by 2 and subtracting from the second row. This is akin to putting the equations in a format where you can more easily see the relationship between the variables, thereby making it easier to extract solutions for each variable in terms of the others. It is essential to use these operations judiciously because each operation must not change the solution set of the system.

Matrix Representation

Matrix representation is a compact and structured way to represent systems of linear equations. Each row of the matrix corresponds to an equation, and each column corresponds to a coefficient of a variable. When matrices are used, the variables don't need to be repeatedly written, simplifying the visual representation of the system.

In our exercise, the system of equations if represented as a matrix looks like this: \

\[\begin{bmatrix}1 & 2 & 3 & 1 & 6 \2 & 4 & 1 & 2 & 2\end{bmatrix}\]

Matrix representation is crucial as it allows for the application of row operations, which are the basis for many methods of solving systems, including Gaussian elimination and Gauss-Jordan elimination.

Dependent and Independent Variables

In the context of a system of linear equations, dependent and independent variables describe the relationship between variables. An independent variable is one that can be freely chosen without restrictions, while a dependent variable's value depends on the values of the independent variables.

In our exercise, we are tasked with identifying these relationships. For example, after manipulation, we find that the value of v is dependent on the value of z: \

\[v = -\frac{1}{2}z\]

This indicates that once a value for z is chosen, the value for v is automatically determined. Understanding the distinction between dependent and independent variables helps in recognizing the direction of dependency in the system, thus making it possible to express some variables in terms of others.

Solving Linear Systems

Solving linear systems involves finding the values of the variables that satisfy all the equations simultaneously. The strategies include graphing the equations to find the point of intersection, substitution, elimination, and using matrix operations such as Gaussian elimination.

The exercise demonstrates solving a linear system by first employing matrix representation and row operations to simplify the system, leading to easier extraction of the relationships between variables. The final step often requires back-substitution where the values of some determined variables are plugged back into previous equations to find the values of the remaining variables.

In this specific case, the process revealed that certain variable pairs, such as (z, v), have a dependent relationship that can be detailed as functions of each other, ultimately leading to a comprehensive solution for the system.