Question

Use back-substitution to solve the system of linear equations. $$\left\\{\begin{aligned} x-y+z &=4 \\ 2 y+z &=-6 \\ z &=-2 \end{aligned}\right.$$

Short answer

The solutions to the system of equations are x = 4, y = -2, and z = -2.

Step-by-step solution

Start with the Given Solution for z

The third equation already gives us the solution for z as -2. We'll start the substitution process with this value.

Substitution into Second equation

Substitute z = -2 into the second equation \[2y+z = -6\] . This simplification becomes \[2y+(-2) = -6\]. Solving for y, we get \(y = -2\).

Substitution into First Equation

Next, substitute z = -2 and y = -2 into the first equation \(x - y + z = 4\). After substituting we get \[x - (-2) + (-2) = 4\] simplifying that we find \(x = 4\)

Key concepts

Back-Substitution

Back-substitution is a method used to solve upper triangular systems of linear equations. This means that the system of equations has already been transformed, often through methods like Gaussian Elimination, into a form where each equation contains one more unknown than the previous equation. The strategy involves working from the bottom of the system upwards.

In simpler terms, you start by solving the last equation in the system, which should have only one unknown. Once you have this solution, you substitute this back into the preceding equations, progressively solving for each of the unknowns.

• Begin with the equation with just one unknown, solve for that unknown.
• Use that value to find the next unknown in the equation above it.
• Continue the process until all unknowns are solved.

In the original exercise, we use back-substitution to solve for the variables starting with equation 3 (from bottom to top):

• First solve for \(z = -2\) from the third equation.
• Substitute \(z = -2\) into the second equation to solve for \(y\).
• Finally, use both \(y = -2\) and \(z = -2\) in the first equation to solve for \(x\).

The benefit of back-substitution is its simplicity once the system is in triangular form.

Linear Algebra

Linear Algebra is a branch of mathematics that deals with vectors, vector spaces, and systems of linear equations. It is fundamental in understanding the nature of lines, planes, and other higher dimensional spaces through algebraic means. Linear algebra provides tools to model and solve problems involving linear relationships.

At the core of linear algebra are matrices and vectors. A system of linear equations, like the one in our exercise, can be represented as a matrix equation. This is a powerful way to organize and manipulate the equations efficiently.

• Vectors represent points or directions in space.
• Matrices are grids of numbers that can represent linear transformations between different vector spaces.
• Understanding matrices is crucial for solving systems of linear equations.

For example, in solving our system via substitution or elimination, one might initially convert the system into matrix form to apply broader linear algebra techniques. This helps to visualize and simplify solving more complex systems as the concept of linear independence and span can be easily analyzed within vector spaces.

Gaussian Elimination

Gaussian Elimination is a systematic method used in linear algebra to solve systems of linear equations. It transforms a given system into a reduced echelon form (upper triangular form) where back-substitution can easily be applied. This method consists of sequential row operations to simplify the system step-by-step.

The goal is to get zeros below the pivot positions, which are the leading coefficients in each row. Once in this form, solving for each variable becomes straightforward.

• Forward Elimination: Use row operations to create zeros below each pivot.
• Pivoting: Rearrange rows to ensure the pivot elements are non-zero and ideally largest in their column.
• Back-Substitution: After achieving the triangular form, solve for variables starting from the last row up to the first.

Though not directly appearing in the step-by-step solution, Gaussian Elimination conceptually simplifies the system into one like in the exercise where back-substitution can take place. It's the groundwork that makes solving even large systems straightforward.