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The systems of linear equations are in upper triangular form. Find all solutions of each system. $$ \left\\{\begin{aligned} 2 x+\quad y+\quad z &=-9 \\ 3 y-2 z &=-4 \\ 8 z &=-8 \end{aligned}\right. $$

Short Answer

Expert verified
The solution to the system is \(x = -3\), \(y = -2\), and \(z = -1\).

Step by step solution

01

Solve the Equation for z

Start with the last equation of the system, which is already in terms of a single variable: \(8z = -8\). Divide both sides by 8 to solve for \(z\): \[ z = \frac{-8}{8} = -1 \].
02

Substitute z into the Second Equation

Now substitute \(z = -1\) into the second equation \(3y - 2z = -4\) and solve for \(y\): \[ 3y - 2(-1) = -4 \]. This simplifies to \(3y + 2 = -4\). Subtract 2 from both sides: \(3y = -6\). Divide by 3: \[ y = \frac{-6}{3} = -2 \].
03

Substitute y and z into the First Equation

Substitute \(y = -2\) and \(z = -1\) into the first equation \(2x + y + z = -9\) and solve for \(x\): \[ 2x - 2 - 1 = -9 \]. Simplifying gives \(2x - 3 = -9\). Add 3 to both sides: \(2x = -6\). Divide by 2: \[ x = \frac{-6}{2} = -3 \].

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Key Concepts

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

Upper Triangular Form
The concept of an upper triangular form in a system of linear equations is crucial for simplifying the process of finding solutions. In this form, the equations are structured so that each subsequent equation contains fewer variables than the previous ones. This structure typically results in a matrix form resembling a triangle when written out.

For example, consider a 3x3 system of equations arranged like trapezoidal stairs:
  • The first equation includes all the variables.
  • The second equation has fewer variables, starting with the second one.
  • The last equation usually contains just one variable.


This makes it easier to solve as you can start with the bottom equation, find the value for that variable, and substitute back up into the previous equations. Solving in this order ensures each step involves only one new unknown at a time.
Solutions of Linear Equations
When solving systems of linear equations, the goal is to find the values of the variables that satisfy all given equations simultaneously. Solutions to these systems can vary:
  • If there is a unique solution, it means there is one particular set of values that satisfies all equations.
  • Infinite solutions can exist if the equations are dependent, meaning they actually represent the same geometric plane.
  • If no solution exists, the equations represent conflicting constraints, indicating they are parallel lines or planes that never intersect.


In the given exercise, the system is structured to ensure a unique solution by using clear, non-redundant equations. Each step simplifies the system, reducing it into clearer relationships, and ultimately, results in a singular intersection point in the solution space.
Step-by-Step Solution
A step-by-step solution approach breaks down the process of solving systems of equations into manageable parts. Tackling the equations in a particular order is key to making the solution process systematic and error-free.

**Step 1: Solve the Simplest Equation**
Start solving from the bottom equation in the upper triangular form, where generally only one variable is involved. This provides a direct solution for one of the variables.

**Step 2: Substitute Back to Solve** Once the value of one variable is determined, substitute this back into the higher equations where that variable appears, reducing them to single-variable equations.

**Step 3: Continue the Substitution Process**
Continue substituting known variable values into previous equations. This method, like peeling layers of an onion, gradually unveils each unknown till you solve for all variables.

This sequential approach ensures each step logically follows the previous, effectively narrowing down the potential unknowns until all variables are determined.

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Most popular questions from this chapter

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{aligned} 4 u+3 v-2 w &=14 \\ u+2 v-3 w &=6 \\ 2 u-v+4 w &=2 \end{aligned}\right.$$

In this exercise, we continue to explore some of the connections between matrices and geometry. As in Exercise \(59,\) we will use \(2 \times 1\) matrices to specify the coordinates of points in the plane. Let \(P, S,\) and \(T\) be the matrices defined as follows: $$ \begin{array}{c} P=\left(\begin{array}{c} \cos x \\ \sin x \end{array}\right) \quad S=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right) \\ T=\left(\begin{array}{cc} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{array}\right) \end{array} $$ Notice that the point \(P\) lies on the unit circle. (a) Compute the matrix \(S P .\) After computing \(S P,\) observe that it represents the point on the unit circle obtained by rotating \(P\) (about the origin) through an angle \(\theta\) (b) Show that $$ S T=T S=\left(\begin{array}{cc} \cos (\theta+\beta) & -\sin (\theta+\beta) \\ \sin (\theta+\beta) & \cos (\theta+\beta) \end{array}\right) $$ (c) Compose (ST)P? What is the angle through which \(P\) is rotated?

The U.S. Food and Drug Administration lists the following RDI's (reference daily intakes) for the antioxidants vita\(\min C,\) vitamin \(E,\) and zinc. Vitamin C: \(60 \mathrm{mg} \quad\) Vitamin \(\mathrm{E}: 30 \mathrm{mg} \quad\) Zinc: \(15 \mathrm{mg}\) Remark on terminology: The Food and Drug Administration defines \(R D I\) as a weighted average of the recommended daily allowances for all segments of the U.S. population. Suppose that you have three brands of dietary supplements on your shelf. Among other ingredients, all three contain the antioxidants mentioned above. The amounts of these antioxidants in each supplement are indicated in the following table. How many ounces of each supplement should you combine to obtain the RDI's for vitamin \(C\), vitamin E, and zinc? $$\begin{array}{lccc} & \begin{array}{c} \text { Vitamin C } \\ \text { (mg/oz.) } \end{array} & \begin{array}{c} \text { Vitamin E } \\ \text { (mg/oz.) } \end{array} & \begin{array}{c} \text { Zinc } \\ \text { (mg/oz.) } \end{array} \\ \hline \text { Supplement I } & 12 & 4 & 1 \\ \text { Supplement II } & 5 & 1.25 & 2.5 \\ \text { Supplement III } & 2 & 3 & 0.5 \\ \hline \end{array}$$

Let \(a, b,\) and \(c\) be constants (with \(a \neq 0\) ), and consider the system $$\left\\{\begin{array}{l}y=a x^{2}+b x+c \\\y=k\end{array}\right.$$ For which value of \(k\) (in terms of \(a, b,\) and \(c\) ) will the system have exactly one solution? What is that solution? What is the relationship between the solution you've found and the graph of \(y=a x^{2}+b x+c ?\)

So the input (3,5) yields an output of \(\sqrt{2} .\) We define the do- main for this function just as we did in Chapter 3: The domain is the set of all inputs that yield real-number outputs. For instance, the ordered pair (1,4) is not in the domain of the function we have been discussing, because (as you should check for yourself\() f(1,4)=\sqrt{-1},\) which is not a real number. We can determine the domain of the function in equation ( 1 ) by requiring that the quantity under the radical sign be non negative. Thus we require that \(2 x-y+1 \geq 0\) and, consequently, \(y \leq 2 x+1\) (Check this.) The following figure shows the graph of this inequality; the domain of our function is the set of ordered pairs making up the graph. In Exercises follow a similar procedure and sketch the domain of the given function. (Graph cant copy) $$g(x, y)=\sqrt{25-x^{2}-y^{2}}$$

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