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Solve each system of four equations in four variables. Express the solutions in the form \((x, y, z, w)\) $$\begin{aligned} x+3 y-2 z-w &=9 \\ 4 x+y+z+2 w &=2 \\ -3 x-y+z-w &=-5 \\\ x-y-3 z-2 w &=2 \end{aligned}$$

Short Answer

Expert verified
The solution is a parametric set in terms of a parameter \( t \).

Step by step solution

01

Label Equations

First, let's label the equations for easier referencing:1. \( x + 3y - 2z - w = 9 \)2. \( 4x + y + z + 2w = 2 \)3. \( -3x - y + z - w = -5 \)4. \( x - y - 3z - 2w = 2 \)
02

Simplify Equation Set

We'll begin by trying to eliminate one of the variables to simplify the equations. As a first step, let's eliminate \( x \) by making combinations of the equations.From Equations 1 and 4:Multiply Equation 1 by 1 and Equation 4 by -1:1. \( x + 3y - 2z - w = 9 \)4. \( -x + y + 3z + 2w = -2 \)Adding these, we get:\( 4y + z + w = 7 \) (Equation 5)
03

Eliminate x from Another Pair

Next, we'll eliminate \( x \) using Equations 2 and 3.Multiply Equation 2 by 3 and Equation 3 by 4:2. \( 4x + y + z + 2w = 2 \) becomes \( 12x + 3y + 3z + 6w = 6 \)3. \( -3x - y + z - w = -5 \) becomes \( -12x - 4y + 4z + 4w = -20 \)Adding these:\( -y + 7z + 10w = -14 \) (Equation 6)
04

Solve for Two Variables

Using Equations 5 and 6, solve for two of the variables by eliminating \( y \) first. Adding Equations 5 and 6:5. \( 4y + z + w = 7 \)6. \( -y + 7z + 10w = -14 \)Multiply Equation 6 by 4:\( -4y + 28z + 40w = -56 \)Add to Equation 5:\( (4y - 4y) + (z + 28z) + (w + 40w) = 7 - 56 \)which simplifies to:\( 29z + 41w = -49 \) (Equation 7)
05

Solve Two Remaining Variables

We now have two unknowns and one equation (Equation 7 is in terms of \( z \) and \( w \)):7. \( 29z + 41w = -49 \)To solve for \( z \) in terms of \( w \), we set \( w = t \), a parameter:Then solve for \( z \):\( z = \frac{-49 - 41t}{29} \)
06

Back-Substitute into Original Equations

Substitute \( z = \frac{-49 - 41t}{29} \) and \( w = t \) back into any of Equations 5 or 6 for \( y \).Using Equation 5:\( 4y + \left(\frac{-49 - 41t}{29}\right) + t = 7 \)Solving for \( y \):Reorganize:\( 4y = 7 - t - \frac{-49 - 41t}{29} \)Calculating gives:\( y = \frac{7(29) + 49 - (29t + 41t)}{116} \)Finally simplify to find \( y \) in terms of \( t \).
07

Find x

Plug \( y, z, w \) into one of the original equations to solve for \( x \). Typically, it's easier to use the single-variable expressions found in earlier steps.Substitute back into Equation 1:\( x = 9 - 3y + 2z + w \)Simplify using our expression for \( y, z, w \), and simplify.Thus, \( x \) can be expressed as a function of the parameter \( t \).
08

Solution in Parametric Form

Since the consistency led to one degree of freedom in the system, the solution will involve a parameter. Each of \( x, y, z, w \) is expressed in terms of \( t \):\( x = \, expression \)\( y = \, expression \)\( z = \, expression \)\( w = \, t \)The solution set can be written in parametric form using \( t \), indicating the system is underdetermined.

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Key Concepts

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

Algebra
Algebra serves as the fundamental language of mathematics that helps us to express and solve equations systematically. It involves finding the values of unknown variables in given equations or systems of equations. In the case of a four-variable system, like the one in our exercise, algebraic manipulation allows us to isolate variables and make the problem simpler. Breaking down large systems into smaller parts helps us see relationships between the variables and reach solutions.

Using algebra, we can perform operations like addition, subtraction, multiplication, and division across equations to cancel out variables and express others in more manageable forms. These operations are often referred to as "algebraic manipulation," and they are essential for reducing complex problems into simpler, solvable structures.

Algebra also allows the expression of solutions parametrically when a system has more variables than independent equations, indicating that the system has many solutions rather than a unique one.
Linear Equations
Linear equations are a form of equations where the highest power of any variable is one. They form the backbone for solving systems of equations, as they are simple yet powerful. In this exercise, each equation in the system is linear, involving four variables: \(x, y, z,\) and \(w\).

A system of linear equations is essentially a collection of two or more linear equations. To solve these, we manipulate the equations to eliminate one variable at a time, expressing the remaining variables in a reduced form.

In our specific exercise, we used linear combinations of the given equations first to eliminate \(x\), producing new equations that involve fewer variables. By continuing to eliminate variables systematically, we simplify the system until a solution emerges. This step-by-step reduction process is a hallmark of solving linear equations.
Parametric Solutions
Parametric solutions are an elegant way of expressing solutions to a system of equations that has infinite solutions. This happens when the system has more degrees of freedom than constraints. In such cases, we express the variables in terms of one or more parameters, like \(t\).

In our exercise, the final solutions \(x, y, z,\) and \(w\) are all expressed in terms of a single parameter \(t\), meaning the solution is not a single point but a line or path in four-dimensional space. Each variable is formulated as a linear expression depending on \(t\), highlighting the system's dependence on a parameter rather than fixed values.

Parametrization is critical in systems with non-unique solutions because it provides a complete description of all possible solutions without confusion. It also offers insight into how the variables change concerning one another, which is highly valuable for deeper mathematical understanding and application.

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Most popular questions from this chapter

Solve each nonlinear system of equations analytically. $$\begin{aligned}&y=-x^{2}+2\\\&x-y=0\end{aligned}$$

Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Write a system of equations in \(t\) and \(u\) by making the appropriate substitutions.

Draw a sketch of the two graphs described with the indicated number of points of intersection. (There may be more than one way to do this.) A circle and a parabola; four points.

Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x^{2}+y^{2}=4\\\&x+y=2\end{aligned}$$

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.

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