Problem 22 - Precalculus

Question

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. $$ \left\\{\begin{aligned} x-y+z &=8 \\ 3 x+3 y-9 z &=-6 \\ 7 x-2 y+5 z &=39 \end{aligned}\right. $$

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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:\[y - 2z = -5\]

Step 3: 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:\[0x + 5y - 2z = -17\]

Step 4: 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.

Step 5: 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 \[z = 1\].

Step 6: 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:\[x + 3 + 1 = 8\]\[x = 8 - 4 = 4\].

Step 7: 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.

Key Concepts

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—the first non-zero element in each row—to 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.