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Chapter 8: Problem 22

In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent, consistent dependent, or inconsistent. $$ \left\\{\begin{aligned} x-y+z &=8 \\ 3 x+3 y-9 z &=-6 \\ 7 x-2 y+5 z &=39 \end{aligned}\right. $$

Short Answer

Expert verified
The system is consistent independent with the solution \((4, -3, 1)\).

Step by step solution

01

Express the System in Triangular Form

To express the system in triangular form, we can use the method of Gaussian elimination. We aim to transform the system so that each equation after the first has fewer variables. Start with the given system: \[\begin{align*}x - y + z &= 8 \3x + 3y - 9z &= -6 \7x - 2y + 5z &= 39\end{align*}\]We need to eliminate the variable \( x \) from the second and third equations. To do this, multiply the first equation by 3 and subtract it from the second equation. For the third equation, multiply the first equation by 7 and subtract it from the third equation.
02

Eliminate x from the Second Equation

Multiply the first equation by 3 to help eliminate \( x \) from the second equation:\[3(x - y + z) = 3 \times 8\]which becomes:\[3x - 3y + 3z = 24\]Subtract this from the second equation:\[3x + 3y - 9z = -6\]Resulting in:\[0x + 6y - 12z = -30\]Simplify this to:\[y - 2z = -5\]
03

Eliminate x from the Third Equation

Multiply the first equation by 7 to help eliminate \( x \) from the third equation:\[7(x - y + z) = 7 \times 8\]which becomes:\[7x - 7y + 7z = 56\]Subtract this from the third equation:\[7x - 2y + 5z = 39\]Resulting in:\[0x + 5y - 2z = -17\]
04

Further Simplify the System

Now we have this triangular form system:\[\begin{align*}x - y + z &= 8 \y - 2z &= -5 \5y - 2z &= -17\end{align*}\]Next, eliminate \( y \) from the third equation by manipulating the equations.
05

Eliminate y from the Third Equation

Multiply the second equation by 5 to get the coefficients of \( y \) in the second and third equations the same:\[5(y - 2z) = 5 \times -5\]This gives:\[5y - 10z = -25\]Subtract this result from the third equation:\[5y - 2z = -17\]Resulting in:\[0y + 8z = 8\] which simplifies to \[z = 1\].
06

Back Substitute to Find y and x

Using \( z = 1 \), substitute back into the second equation:\[y - 2(1) = -5\]\[y - 2 = -5\]\[y = -3\]Finally, substitute \( y = -3 \) and \( z = 1 \) into the first equation:\[x + 3 + 1 = 8\]\[x = 8 - 4 = 4\].
07

Classification of the System

The solution found is \((x, y, z) = (4, -3, 1)\), which is a single solution. Thus, the system is **consistent independent**, as it has a unique solution.

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Key Concepts

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

Gaussian elimination
Gaussian elimination is a fundamental technique used to solve systems of linear equations. The method consists of two main stages: transforming the coefficient matrix into a certain form and then solving the system through back substitution. This process makes it easier to handle the equations and is widely used because it can solve any system of linear equations that has a solution.

Key Steps to Perform Gaussian Elimination:
  • Choose a pivot element鈥攖he first non-zero element in each row鈥攖o work with. This is the element you'll use to eliminate variables from other equations.
  • Use row operations to create zeros below the pivot element. Do this by multiplying the pivot equation by a suitable number, then subtract it from another equation to eliminate the chosen variable from that row.
  • Continue the process for all pivot elements in the system.
By executing these steps systematically, the system of equations is transformed into simpler and more manageable forms, ultimately reaching what is known as triangular form.
Triangular form
The concept of triangular form is crucial in simplifying and solving systems of linear equations. When a system is in triangular form, each equation resembles a staircase: every row has fewer variables than the previous row.

Characteristics of Triangular Form:
  • The first equation has the most variables, and as you progress down the list, each subsequent equation has one less variable.
  • Typically, the bottom row has one variable. It provides a starting point for back substitution to solve the system.
  • This arrangement allows for systematic solution methods, making it straightforward to substitute back through into the previous equations to find all values.
Triangular form is an intermediary step in Gaussian elimination, allowing easier computation and paving the way for back substitution.
Consistent independent
In the world of systems of linear equations, a consistent independent system is one that has exactly one unique solution. This means that all lines or planes represented by the equations intersect precisely at one point.

Indicators of a Consistent Independent System:
  • The row-reduced form of the augmented matrix does not contain any row entirely consisting of zeros followed by a non-zero number, which would indicate a contradiction.
  • There are as many pivot columns as there are variables, ensuring each variable can be solved uniquely.
  • A graphical representation shows all equations meeting at a single intersection point.
This type of system is ideal for many practical applications where a definitive solution is needed, as it provides a clear and definitive answer to the problem.
Back substitution
Back substitution is the final step in solving a system of linear equations that has been reduced to triangular form. It involves finding the values of the unknowns starting from the last equation and moving upwards.

Steps in Back Substitution:
  • Begin with the last equation that typically has only one variable, solve for this variable.
  • Substitute this value into the previous equation to solve for another variable.
  • Continue the process upwards until all variables have been found.
For example, if the reduced system gives you equations:\[\begin{align*} y - 2z &= -5 \ z &= 1 \end{align*} \] start by solving for \( z \), how we found \( z = 1 \). Substitute back to find \( y = -3 \), then use these values in the first equation to solve for \( x \).

Back substitution efficiently reveals the solution to each variable in the original system of equations.

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Most popular questions from this chapter

Use Cramer鈥檚 Rule to solve the system of linear equations. $$ \left\\{\begin{aligned} x+y &=8000 \\ 0.03 x+0.05 y &=250 \end{aligned}\right. $$

Consider the following definitions. A square matrix is said to be an upper triangular matrix if all of its entries below the main diagonal are zero and it is said to be a lower triangular matrix if all of its entries above the main diagonal are zero. For example,$$ E=\left[\begin{array}{rrr} 1 & 2 & 3 \\\ 0 & 4 & -9 \\ 0 & 0 & -5 \end{array}\right] $$ from Exercises 8 - 21 above is an upper triangular matrix whereas $$ F=\left[\begin{array}{ll} 1 & 0 \\\ 3 & 0 \end{array}\right] $$ is a lower triangular matrix. (Zeros are allowed on the main diagonal.) Discuss the following questions with your classmates. Is the product of two \(n \times n\) upper triangular matrices always upper triangular?

Solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=25 \\ 4 x^{2}-9 y &=0 \\ 3 y^{2}-16 x &=0 \end{aligned}\right. $$

Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to find the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables \(x, y\) and \(z\) are the dimensions of the box and \(\lambda\) is called a Lagrange multiplier. With the help of your classmates, solve the system. $$ \left\\{\begin{aligned} 2 y+2 z &=\lambda y z \\ 2 x+2 z &=\lambda x z \\ 2 y+2 x &=\lambda x y \\ x y z &=1000 \end{aligned}\right. $$

Sketch the solution to each system of nonlinear inequalities in the plane. $$ \left\\{\begin{aligned} y>& 10 x-x^{2} \\ y &
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