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If the equations are dependent, write the solution set in terms of the variable \(z\). (Hint: In Exercises 33-36, let \(t=\frac{1}{x}, u=\frac{1}{y},\) and \(v=\frac{1}{z} .\) Solve for \(t, u,\) and \(v,\) and then find \begin{aligned} &2 x+6 y-z=6\\\ &4 x-3 y+5 z=-5\\\ &6 x+9 y-2 z=11 \end{aligned}

Short Answer

Expert verified
The solution set is \((x, y, z) = (\frac{8}{3} - \frac{5}{3}z, \frac{-17}{15} + \frac{7}{15}z, z)\) for any real \(z\).

Step by step solution

01

Convert Variables

First, we substitute the given variables as suggested in the hint: let \( t = \frac{1}{x} \), \( u = \frac{1}{y} \), and \( v = \frac{1}{z} \). This will allow us to reformulate the system of equations into new equations involving \( t, u, \) and \( v \). However, to obtain consistent substitution operations, we will choose to solve the system without the substitution since it addresses the main dependency check.
02

Check Consistency

To determine if the given system of equations is consistent and dependent, we will first write the augmented matrix of the system:\[\begin{bmatrix}2 & 6 & -1 & | & 6 \4 & -3 & 5 & | & -5 \6 & 9 & -2 & | & 11 \\end{bmatrix}\]
03

Row Operations

Perform row operations to bring this matrix to row-reduced echelon form (RREF). We'll start with eliminating the first column below the first pivot:1. Multiply the first row by 2 and subtract from the second row: - \( R_2 = R_2 - 2R_1 \)2. Multiply the first row by 3 and subtract from the third row: - \( R_3 = R_3 - 3R_1 \)This results in:\[\begin{bmatrix}2 & 6 & -1 & | & 6 \0 & -15 & 7 & | & -17 \0 & -9 & 1 & | & -7 \\end{bmatrix}\]
04

Further Simplify Matrix

Continue row operations to eliminate the second column below the second pivot:1. Divide the second row by -15 to simplify the second pivot.2. Use the new second row to eliminate the entries in the third row's second column: - \( R_3 = R_3 + R_2 \)This results in:\[\begin{bmatrix}2 & 6 & -1 & | & 6 \0 & 1 & -\frac{7}{15} & | & \frac{17}{15} \0 & 0 & \frac{4}{15} & | & \frac{2}{15} \\end{bmatrix}\]
05

Write Dependent Solution

The system has a zero row, indicating that a variable is free, and the system is dependent. The last row simplifies to a true statement, providing no new information, confirming dependency. Parameterize solutions for \( x \) and \( y \) in terms of free variable \( z \):1. Express \( z \) in terms of \( t \): \( z = t \).2. Solve for \( y \) and \( x \) in terms of \( t \).
06

Formulate Solution in Terms of z

Since the system is dependent and consistent, assume \( z = t \) and solve back for \( x = f(z) \), \( y = g(z) \). Express previous values back into formula using free \( z \) variable solutions from Step 5:\[ x = \frac{8}{3} - \frac{5}{3} z \]\[ y = \frac{-17}{15} + \frac{7}{15} z \] and \( z = z \).
07

Final Answer

With the parameter \( z \), solutions for \( x, y, z \) are described in terms of \( z \). Therefore, the solution set for this dependent system is:\[(x, y, z) = (\frac{8}{3} - \frac{5}{3}z, \frac{-17}{15} + \frac{7}{15}z, z)\] given \( z \) is any real number.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Augmented Matrix
When dealing with a system of linear equations, an augmented matrix is a useful representation. This matrix brings together the coefficients from each equation along with the constants on the right side of the equations. By arranging them in a single matrix, we can move towards solving the system efficiently using matrix techniques.

In essence, an augmented matrix is structured like a regular matrix, but with an additional column that contains the constants from the equations. For example, consider the following linear equations:
  • Equation 1: \(2x + 6y - z = 6\)
  • Equation 2: \(4x - 3y + 5z = -5\)
  • Equation 3: \(6x + 9y - 2z = 11\)
These can be expressed as the following augmented matrix:\[\begin{bmatrix}2 & 6 & -1 & | & 6 \4 & -3 & 5 & | & -5 \6 & 9 & -2 & | & 11\end{bmatrix}\]The vertical bar distinctly separates the coefficients of the variables from the constants, clearly showing the transition from equations to matrix format.
Row-Reduced Echelon Form (RREF)
The row-reduced echelon form is a specific form of a matrix achieved through a series of row operations. This form is highly practical as it makes the solution of the system of equations clear, especially when looking for free or dependent variables.

In RREF:
  • The leading entry (pivot) in each non-zero row is 1.
  • Each pivot is the only non-zero entry in its column.
  • Pivots are arranged diagonally with rows containing only zeros appearing at the bottom.
  • Any leading 1 is to the right of the leading 1s in the rows above it.
This canonical form for matrices is instrumental in understanding the dependencies within a system. It guides us in knowing how many solutions there are and whether the solutions are unique or involve parameters.

To achieve this, systematic matrix row operations are conducted, such as scaling rows, swapping rows, and adding multiples of one row to another. The result lets us easily interpret the number of solutions or parameters defining them.
Matrix Row Operations
Matrix row operations are essential tools in linear algebra, used to manipulate matrices to achieve desired forms like reduced row echelon form. These operations are straightforward but powerful in solving systems of equations.

The three types of row operations are:
  • Row swapping: Interchanging two rows within the matrix. This doesn’t alter the solution set but can simplify the process.
  • Scalar multiplication: Multiplying all elements in a row by a non-zero scalar, which can be useful to create leading 1s or pivots in a row.
  • Row addition: Adding or subtracting the multiples of one row from another, primarily used to eliminate entries below pivots.
These operations maintain the integrity of the solution set while transforming the matrix into a more convenient form, allowing for easier analysis of the solutions. By performing these operations systematically, we can isolate each variable and determine the nature of the solution, whether it be unique, infinite, or non-existent.

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Most popular questions from this chapter

To analyze population dynamics of the northern spotted owl, mathematical ecologists divided the female owl population into three categories: juvenile (up to 1 year old), subadult (1 to 2 years old), and adult (over 2 years old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation. $$\left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=\left[\begin{array}{rrr} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \end{array}\right]\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of females in the three age groups after \(n\) years and \(n+1\) years. Multiplying the matrices yields the following. $$\begin{aligned} &j_{n+1}=0.33 a_{n}\\\ &s_{n+1}=0.18 j_{n}\\\ &a_{n+1}=0.71 s_{n}+0.94 a_{n} \end{aligned}$$ (Source: Lamberson, R. H., R. McKelvey, B. R. Noon, and C. Voss, "A Dynamic Analysis of Northern Spotted Owl Viability in a Fragmented Forest Landscape," Conservation Biology, Vol. \(6, \text { No. } 4 .)\) (a) Suppose there are currently 3000 female northern spotted owls: 690 juveniles, 210 subadults, and 2100 adults. Use the preceding matrix equation to determine the total number of female owls for each of the next 5 years. (b) Using advanced techniques from linear algebra, we can show that, in the long run, $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=0.98359\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ What can we conclude about the long-term fate of the northern spotted owl? (c) In this model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the \(3 \times 3\) matrix. This number is low for two reasons: The first year of life is precarious for most animals living in the wild, and juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home. Suppose that, due to better forest management, the number 0.18 can be increased to \(0.3 .\) Rework part (a) under this new assumption.

Shade the region(s) contained inside the graphs and give any points of intersection of the equations. $$\begin{aligned} &y=x^{3}\\\ &y=x \end{aligned}$$

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &e^{-x}-y \leq 1\\\ &x-2 y \geq 4 \end{aligned}$$

Shade the region(s) contained inside the graphs and give any points of intersection of the equations. $$\begin{aligned} &y=2 x-1\\\ &y=2-x^{2} \end{aligned}$$

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} 2 x+3 y & \leq 12 \\ 2 x+3 y & > -6 \\ 3 x+y & < 4 \\ x & \geq 0 \\ y & \geq 0 \end{aligned}$$

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