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Chapter 10: Problem 9

Use back-substitution to solve the triangular system. $$\left\\{\begin{aligned} x+2 y+z &=7 \\ -y+3 z &=9 \\ 2 z &=6 \end{aligned}\right.$$

Short Answer

Expert verified
The solution is \(x = 4\), \(y = 0\), \(z = 3\).

Step by step solution

01

Solve the Last Equation for z

The system of equations is already in triangular form. We start by solving the last equation. The equation is \(2z = 6\). To find \(z\), divide both sides by 2: \(z = \frac{6}{2} = 3\).
02

Substitute z into the Second Equation

Next, we substitute \(z = 3\) into the second equation \(-y + 3z = 9\). This becomes \(-y + 3(3) = 9\). Simplifying gives \(-y + 9 = 9\). To isolate \(y\), subtract 9 from both sides: \(-y = 0\). Therefore, \(y = 0\).
03

Substitute y and z into the First Equation

Now, substitute \(y = 0\) and \(z = 3\) into the first equation \(x + 2y + z = 7\). This becomes \(x + 2(0) + 3 = 7\). Simplify to \(x + 3 = 7\). Subtract 3 from both sides to solve for \(x\): \(x = 4\).
04

Verify the Solution

To ensure the solution is correct, substitute \(x = 4\), \(y = 0\), and \(z = 3\) back into all three original equations. Each equation holds true: \(4 + 2(0) + 3 = 7\), \(-0 + 9 = 9\), and \(2(3) = 6\). Therefore, the solution is verified.

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Key Concepts

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

Triangular System
In mathematics, a **triangular system** refers to a set of linear equations where variables are ordered in such a manner that each equation contains only one new variable as compared to the previous one. Such a system generally aims to simplify computations.

The structure often forms either an upper or lower triangle within a matrix. Here鈥檚 what it means:
  • For an **upper triangular** system, equations are ordered from most variables to least. Each starts with a new variable which is then solved and plugged into preceding equations.
  • In the given exercise, we see a form of an upper triangular system with three equations and three unknowns: x, y, and z. We solve from the bottom equation up, starting with z, then y, and finally x.
By arranging systems into triangular form, solving becomes more straightforward, particularly suited for back-substitution.
Linear Equations
A **linear equation** is an algebraic expression in which each term is either a constant or the product of a constant and a single variable. These equations graphically represent straight lines. Each equation in our given system is linear.

Linear equations often appear in multiple formats, such as:
  • The standard form: Ax + By = C, where A, B, and C are constants.
  • Equations with multiple variables, such as our example, still maintain linearity provided all terms are either constants or first-power variables without exponents.
  • Such equations can represent real-world situations like finance models, physics problems, etc.
In context, the system we solved forms a set of linear equations essential for back-substitution techniques.
System of Equations
A **system of equations** consists of multiple equations used together to determine the values of shared variables. Each system can be either independent, dependent, or inconsistent based on its solutions.

Solving involves finding such a combination of values for variables that satisfy all equations simultaneously. This might entail various methodologies, like graphical solving, substitution, or elimination. In our task:
  • We used a pre-arranged triangular form for convenience.
  • Back-substitution is most efficient when handling systems in triangular forms, fostering easier solving by directly attacking for variable values.
The main aim here is recognizing a compatible solution set fulfilling all equations simultaneously.
Step-by-Step Solution
A **step-by-step solution** provides a detailed walk-through on tackling complex problems, breaking them into manageable parts. This method ensures understanding is built progressively.

In our solution for the original exercise:
  • First, we cracked the simplest solution from the last equation and moved upwards.
  • This approach leverages clarity by sequentially isolating each variable, fixing it before progressing.
  • Step-by-step guides are invaluable for learning due to their logical flow and easy remembrance.
Such methodology anchors the learning and understanding of problem-solving, making it conducive for educational contexts and easier comprehension for students.

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Most popular questions from this chapter

Find the inverse of the matrix if it exists. $$\left[\begin{array}{llll} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{array}\right]$$

Powers of a Matrix Let $$ A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] $$ Calculate \(A^{2}, A^{3}, A^{4}, \ldots\) until you detect a pattern. Write a general formula for \(A^{n}\)

An encyclopedia saleswoman works for a company that offers three different grades of bindings for its encyclopedias: standard, deluxe, and leather. For each set that she sells, she earns a commission based on the set's binding grade. One week she sells one standard, one deluxe, and two leather sets and makes \(\$ 675\) in commission. The next week she sells two standard, one deluxe, and one leather set for a \(\$ 600\) commission. The third week she sells one standard, two deluxe, and one leather set, earning \(\$ 625\) in commission. (a) Let \(x, y,\) and \(z\) represent the commission she earns on standard, deluxe, and leather sets, respectively. Translate the given information into a system of equations in \(x, y\) and \(z\) (b) Express the system of equations you found in part (a) as a matrix equation of the form \(A X=B\). (c) Find the inverse of the coefficient matrix \(A\) and use it to solve the matrix equation in part (b). How much commission does the saleswoman earn on a set of encyclopedias in each grade of binding?

Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{array}{l} y \geq x^{2} \\ y \leq 4 \\ x \geq 0 \end{array}\right.$$

Use a calculator that can perform matrix operations to solve the system, as in Example 7 . $$\left\\{\begin{array}{l} 3 x+4 y-z=2 \\ 2 x-3 y+z=-5 \\ 5 x-2 y+2 z=-3 \end{array}\right.$$
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