Problem 36 - Algebra and Trigonometry

Question

35–46 Solve the system of linear equations. $$\left\\{\begin{array}{l}{2 x-3 y+5 z=14} \\ {4 x-y-2 z=-17} \\\ {-x-y+z=3}\end{array}\right.$$

Step-by-Step Solution

Answer

The system has infinite solutions. General solution: \((x, y, z) = (32 - 11t, t, 2.5t - 2.5), \ t \in \mathbb{R}\).

Step 1: Eliminate the x variable

To start solving the system, add equation 1 and 3 to eliminate \(x\). This gives:\[(2x - 3y + 5z) + (-x - y + z) = 14 + 3\]Simplifying, we get:\[x - 4y + 6z = 17\]

Step 2: Eliminate the x variable again

Next, we'll eliminate \(x\) from equations 2 and 3. Multiply equation 3 by 4:\[-4x - 4y + 4z = 12\]Add this to equation 2:\[(4x - y - 2z) + (-4x - 4y + 4z) = -17 + 12\]Simplifying, we get:\[-5y + 2z = -5\]

Step 3: Solve the new system of equations

Now, we solve the new system:1. \(x - 4y + 6z = 17\) 2. \(-5y + 2z = -5\)Multiply the second equation by 2 to help eliminate \(y\):\[-10y + 4z = -10\]Substitute this into equation 1 by isolating \(z\) from modified equation 2:\[4z = 10y - 10\]\[z = 2.5y - 2.5\]

Step 4: Solve for y and z

Substitute \(z = 2.5y - 2.5\) into the second equation: \[-5y + 2(2.5y - 2.5) = -5\]Simplifying gives:\[-5y + 5y - 5 = -5\]This simplifies to 0 = 0, indicating infinite solutions for \(y\) when solved further uniquely.

Step 5: Solve for x

To find specific solutions, plug \(z = 2.5y - 2.5\) back into the equation:\[x - 4y + 6(2.5y - 2.5) = 17\]Simplifying:\[x - 4y + 15y - 15 = 17\]\[x + 11y = 32\]This means for any value of \(y\), \(x = 32 - 11y\).

Step 6: General solution for the system

From Step 5, we identified parameter forms for \(x\), \(y\), and \(z\):\[x = 32 - 11t\]\[y = t\]\[z = 2.5t - 2.5\]where \(t\) is any real number. This parameterization represents the infinite solutions for the system.

Key Concepts

Elimination Method

The elimination method is a powerful technique for solving systems of linear equations by systematically removing variables. This is accomplished by adding or subtracting equations to cancel out a particular variable, making the system easier to solve.
For example, when solving the given system:

• We first aim to eliminate the variable \(x\).
• Add the first and the third equations: \(2x - 3y + 5z = 14\) and \(-x - y + z = 3\).
• Combine them to get: \(x - 4y + 6z = 17\).

By repeating this process on different pairs of equations, we can eliminate variables step-by-step.
This method simplifies a multi-equation problem into one that is more manageable, eventually reducing the system to a form where back substitution can be easily used to find solutions.

Parametric Solutions

Once variables are eliminated and simplified equations are formed, parametric solutions can be identified. These solutions introduce one or more parameters, which can represent an infinite set of solutions. In the exercise, the system of equations was simplified to find relations between variables:

• \(x = 32 - 11t\)
• \(y = t\)
• \(z = 2.5t - 2.5\)

Here, \(t\) is a parameter, representing any real number.
Such parametric forms enable expressing all variables in terms of a single free variable.
This effectively provides a way to describe all possible solutions to the original system of equations in a compact and generalized manner. Parametric solutions are especially useful when a system has infinitely many solutions, as they provide a way to comprehend and express the entire solution set.

Infinite Solutions

The concept of infinite solutions appears when a system of linear equations has more than one solution, potentially infinite. This occurs when the equations describe planes or lines that intersect at multiple points.
In the exercise, during Step 4, refining the equation for \(z\) led to a simplification ending with \(0=0\). This indicates that the original equations are dependent and the system has infinite solutions.

• When we have something like \(0=0\), it implies the equations are not distinct but rather are compatible at many points.
• In the parametric form, the solution space is represented using a parameter, covering every possible solution fitting the equations.

Recognizing infinite solutions allows for a deeper understanding of the geometry of the solutions, seeing how they cover an entire line or plane in multidimensional space.

Linear Algebra Concepts

Solving systems of linear equations touches on key linear algebra concepts which are critical for structuring and solving these mathematical models.
Firstly, linear equations can be thought of in terms of matrices and vectors, with operations involving row reduction or transformation being akin to elimination methods.

• Matrix representation simplifies complex systems.
• It allows for the application of algorithms and software to find solutions efficiently.

Secondly, understanding solution sets in terms of parametric equations relates directly to the concept of vector spaces and dimensionality:

• Parameters describe basis vectors of solution spaces.
• Infinite solutions suggest the span of vectors, illustrating the dimensions of the solution space.

Linear algebra concepts provide the foundational tools needed for analyzing the relationships between equations and finding broader solutions to complex problems.