Problem 37 - Precalculus: Mathematical for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{array}{rr} x+2 y-3 z= & -5 \\ -2 x-4 y-6 z= & 10 \\ 3 x+7 y-2 z= & -13 \end{array}\right.$$

Step-by-Step Solution

Answer

Infinitely many solutions parameterized by z.

Step 1: Analyze the system

Examine the given system of equations: 1) \( x + 2y - 3z = -5 \) 2) \( -2x - 4y - 6z = 10 \) 3) \( 3x + 7y - 2z = -13 \). Notice that equation 2 is simply -2 times equation 1, indicating the system might have no solution or infinitely many solutions.

Step 2: Eliminate variable using substitution

Choose to eliminate variable \(x\). From equation (1), express \(x\) in terms of \(y\) and \(z\): \( x = -5 - 2y + 3z \).

Step 3: Substitute into equation 3

Substitute \( x = -5 - 2y + 3z \) into equation (3): \( 3(-5 - 2y + 3z) + 7y - 2z = -13 \). Simplify to obtain a new equation in terms of \(y\) and \(z\).

Step 4: Simplify the new equation

Simplifying the equation from Step 3: \(-15 - 6y + 9z + 7y - 2z = -13 \) which becomes \( y + 7z = 2 \). This is equation (4).

Step 5: Solve the reduced system

Now we have two key equations considering the contradiction found earlier:1) \( x + 2y - 3z = -5 \) 2) (redundant)3) \( y + 7z = 2 \).From equation (4), express \( y = 2 - 7z \) and substitute in equation 1.

Step 6: Find expressions for variables

Substitute \( y = 2 - 7z \) into equation 1:\( x + 2(2 - 7z) - 3z = -5 \) >\( x + 4 - 14z - 3z = -5 \),\( x - 17z = -9 \), hence \( x = 17z - 9 \).

Step 7: Analyze solution

From equations obtained:\( x = 17z - 9 \),\( y = 2 - 7z \),\( z \) is free.The system has infinitely many solutions, parameterized by \(z\).

Step 8: Express final solution

The solution is parameterized by \(z\), \( x = 17z - 9 \), \( y = 2 - 7z \), \( z = z \), where \( z \) is any real number.

Key Concepts

Solution Methods

Solving systems of linear equations involves finding values for the variables that satisfy all given equations. The methods to solve these systems include substitution, elimination, and using matrices. Each method can be more suitable depending on the system.

• Substitution: Solve one equation for a single variable and substitute it in the other equations. This method is helpful when an equation is easily solvable for a variable.
• Elimination: Add or subtract equations to eliminate one variable, simplifying the system. It is powerful when dealing with linear equations where coefficients line up nicely.
• Matrices: Utilize row operations or apply matrix inversions. This method is effective for larger systems.

For the given problem, substitution was chosen as it allowed immediate simplification by expressing one variable in terms of others.

Infinitely Many Solutions

A system of linear equations may not always have a unique solution. Sometimes, it can have infinitely many solutions or no solution at all. To identify such cases:

• Infinitely Many Solutions: This occurs when the equations describe the same plane or line, leading to overlapping solutions in a system. Essentially, the system is dependent.
• No Solution: This scenario happens when the equations represent parallel planes or lines that never intersect, making the system inconsistent.

In our problem, the presence of infinitely many solutions was indicated by the redundancy of one equation, showing dependency within the equations.

Linear Algebra

Linear algebra plays a crucial role in understanding systems of linear equations and how to solve them.

• Variables and Equations: The concept of variables and how they relate through linear equations is foundational in linear algebra.
• Vectors and Matrices: These tools allow representation of linear systems more compactly and are particularly useful in large and complex systems.
• Linear Independence and Dependence: Understanding these concepts is vital when determining the nature of solutions, such as unique solutions, infinitely many, or none.

In our exercise, linear algebra concepts like equation redundancy and parameterization (expressing solutions in terms of free variables) were used to identify the nature of solutions.