Question

Solve the given set of equations, or else show that there is no solution. $$ \begin{aligned} x_{1} -x_{3}=0 \\ 3 x_{1}+x_{2}+x_{3} &=1 \\\\-x_{1}+x_{2}+2 x_{3} &=2 \end{aligned} $$

Short answer

Question: What is the solution to the system of linear equations given by: $$ \begin{aligned} x_1 - x_3 &= 0 \\ 3x_1 + x_2 + x_3 &= 1 \\ -x_1 + x_2 + 2x_3 &= 2 \end{aligned} $$ Answer: The solution to this system of equations is given by the ordered triple \((x_1, x_2, x_3)\): $$(-\frac{1}{3}, \frac{5}{3}, -\frac{1}{3})$$

Step-by-step solution

Solve the first equation for \(x_1\)

The first equation is given by: $$x_1 - x_3 = 0$$ Adding \(x_3\) to both sides, we get: $$x_1 = x_3$$

Substitute the expression for \(x_1\) in the other two equations

Replacing \(x_1\) with \(x_3\) in the second and third equations, we have: $$ \begin{aligned} 3x_3 + x_2 + x_3 &= 1 \\ -x_3 + x_2 + 2x_3 &= 2 \end{aligned} $$ Now, we can simplify these equations: $$ \begin{aligned} 4x_3 + x_2 &= 1 \\ x_3 + x_2 &= 2 \end{aligned} $$

Solve the second equation for \(x_2\)

Solving the second equation for \(x_2\): $$x_2 = 2 - x_3$$

Substitute the expression of \(x_2\) into the first simplified equation

Replacing \(x_2\) with \(2 - x_3\) in the first simplified equation: $$4x_3 + (2 - x_3) = 1$$ Simplifying this equation, we get: $$3x_3 = -1$$ Dividing both sides by 3, we get: $$x_3 = -\frac{1}{3}$$

Find the value of \(x_2\)

Substituting the value of \(x_3\) into the expression for \(x_2\): $$x_2 = 2 - (-\frac{1}{3})$$ $$x_2 = \frac{5}{3}$$

Find the value of \(x_1\)

Since \(x_1 = x_3\), we have: $$x_1 = -\frac{1}{3}$$

Write the solution as an ordered triple

The solution to this system of equations is given by the ordered triple \((x_1, x_2, x_3)\): $$(-\frac{1}{3}, \frac{5}{3}, -\frac{1}{3})$$

Key concepts

Matrix Algebra

Matrix algebra is a powerful tool for solving systems of linear equations like the ones given in the exercise. Instead of dealing with equations directly, you can represent the system in a matrix format. This involves creating two matrices: the coefficient matrix and the constant matrix. The coefficient matrix contains the coefficients of the variables, whereas the constant matrix includes the constants from the right-hand side of the equations.

Using matrices helps simplify operations such as addition, subtraction, and multiplication of equations. This is because you can use matrix operations to row reduce the system and make it easier to isolate variables. This method is especially useful when dealing with larger systems, as it streamlines calculations and minimizes errors.

Matrix algebra is essential in linear transformations, finding inverses, and determining determinants of matrices. These properties can tell us more about the system, such as its solvability and the type of solutions it may have.

Gaussian Elimination

Gaussian elimination is a systematic method used to solve systems of equations. The goal is to use row operations to transform the coefficient matrix into an upper triangular form. Once in this form, you can use back substitution to find the values of the variables.

In practice, you perform three types of row operations:

• Swap two rows
• Multiply a row by a non-zero scalar
• Add or subtract a multiple of one row to another

These operations aim to simplify the system progressively, by zeroing out the coefficients below the main diagonal in the augmented matrix. In our exercise, Gaussian elimination can be seen through substituting and simplifying equations step by step.

This method is essential because it provides a structured approach to finding solutions or proving that no solution exists, as well as helping determine if the system is consistent or inconsistent.

Consistent System

A consistent system is one in which there is at least one solution. Systems of linear equations can be consistent or inconsistent. When we have a consistent system, as demonstrated in the provided solution, the set of equations intersects at least at one point in the coordinate space.

In the exercise, we found that the solution to the set of equations is \(x_1 = -\frac{1}{3}, x_2 = \frac{5}{3}, x_3 = -\frac{1}{3}\), confirming that the system is consistent. This means that substituting these values back into the original equations yields true statements, satisfying each equation.

A system might have exactly one solution (as in this exercise), infinitely many solutions, or no solution at all if it is inconsistent. Determining the consistency of a system is crucial in understanding its behavior and the nature of its solutions.

Linear Independence

Linear independence is a key concept when analyzing systems of equations. A set of vectors (or equations) is said to be linearly independent if no vector in the set is a linear combination of the others.

For our exercise's system of equations, understanding linear independence helps determine if the solutions are trivial or non-trivial. If the vectors are linearly independent, the only solution to the homogeneous system is the trivial solution \(x_1 = 0, x_2 = 0, x_3 = 0\).

However, for the original (inhomogeneous) system we solved, linear independence indicates that the equations provide unique information about the variables. The given system's equations are not multiples of each other, showing they are independent. This contributes to the solution being unique and hence consistent. In general, determining the linear independence of equations helps understand how information about variables is structured.