# Ideal answer to – how do you know if a linear algebra system is solvable?

Contents

A linear algebra system is solvable if and only if the rank of the augmented matrix is equal to the rank of the coefficient matrix.

Further information is provided below

According to the fundamental theorem of linear algebra, a linear system is solvable if and only if the rank of the augmented matrix is equal to the rank of the coefficient matrix. In other words, if the number of linearly independent rows in the augmented matrix is the same as the number of linearly independent rows in the coefficient matrix, the system is solvable.

A famous quote on the topic comes from mathematician Augustus De Morgan, who said, “The science of pure mathematics, in its modern development, may claim to be the most original creation of the human spirit.” Linear algebra is a prime example of the creative power of mathematics, as it provides tools for solving systems of linear equations that arise in a wide range of applications.

Some interesting facts about linear algebra and the solvability of systems include:

• Linear algebra is used extensively in fields such as physics, engineering, economics, computer science, and many others.
• The rank of a matrix is related to its determinant, which is a scalar value that encodes information about the matrix’s invertibility. If the determinant is zero, the matrix is non-invertible and the system is not solvable.
• If a system is solvable, it may have either one unique solution, infinitely many solutions, or no solutions at all. These possibilities depend on the geometry of the underlying linear space.
• There are many algorithms for solving linear systems, including Gaussian elimination, LU decomposition, QR decomposition, and iterative methods such as Jacobi and Gauss-Seidel. These methods have different strengths and weaknesses and are suited for different types of problems.

To further illustrate the concept of solvability, consider the following example:

\begin{align}
3x + 2y &= 5 \
2x – y &= 4 \
\end{align
}

We can rewrite this system as the augmented matrix

$$\begin{bmatrix} 3 & 2 & 5 \ 2 & -1 & 4 \ \end{bmatrix}$$

Using Gaussian elimination, we can reduce this matrix to row echelon form:

$$\begin{bmatrix} 3 & 2 & 5 \ 0 & -\frac{5}{3} & \frac{2}{3} \ \end{bmatrix}$$

Since there are two linearly independent rows, the rank of the matrix is 2. Therefore, the system is solvable. To find the solution, we can continue with back-substitution or use other methods such as matrix inversion or Cramer’s rule.

IT\\\'S IMPORTANT:  General problems - what jobs can a math grad do?

In summary, the solvability of a linear algebra system depends on the relationship between the rank of the augmented matrix and the rank of the coefficient matrix. The concepts of linear independence, determinants, and matrix inversion are all closely related to this topic, and there are many algorithms for solving linear systems. As De Morgan suggested, linear algebra is a fascinating and creative field of mathematics that has far-reaching applications in many areas of science and technology.

A table summarizing the three possibilities for the solution of a linear system would be:

Number of Solutions Geometry of the Linear Space Rank of the Augmented Matrix
One unique solution The linear space is a point. Number of variables
Infinitely many solutions The equations represent a line, plane, or higher-dimensional subspace. Number of variables < Rank of the Augmented Matrix
No solutions The equations represent parallel lines, planes, or subspaces that do not intersect. Rank of the Coefficient Matrix < Rank of the Augmented Matrix

## Associated video

The video discusses the concept of a system of linear equations being unsolvable and how it can occur in situations where not all necessary information is provided. A specific example of calculating the lengths of five highways is used to show how the lack of solvability can lead to infinitely many solutions or no solutions at all. The importance of ensuring solvability to determine all necessary values is emphasized.

## I discovered more solutions online

• Yes: by showing that the system is equivalent to one in which the equation 0=3 must hold, you have shown the original system has no solutions.
• By definition, a system of linear equation is said to be “consistent” if and only if it has at least one solution; and it is “inconsistent” if and only if it has no solutions. So “showing a system of linear equations is not solvable” (has no solutions) is, by definition, the same thing as showing that the system of linear equations is “inconsistent”.
• “A system doesn’t have a unique solution” can happen in two ways: it can have more than one solution (in which case it has infinitely many solutions), or it can have no solutions. Only in the second case do we say the system is “inconsistent”.
• One of the easiest ways to find solutions of systems of linear equations (or show no solutions exist) is Gauss (or Gauss-Jordan) Row Reduction; it amounts to doing the kind of things you did, but in a systematic, algorithmic, recipe-like manner. You ca…

## Moreover, people are interested

How do you know if a system of equations can be solved?
Response: Compare the m- and b-values of the equations to determine the number of solutions. If the two equations have different m-values, then the system has one solution. If the two equations have the same m-value but different b-values, then the system has no solution.
What is an unsolvable system of linear equations?
The reply will be: An underdetermined linear system has either no solution or infinitely many solutions. For example, is consistent and has an infinitude of solutions, such as (x, y, z) = (1, −2, 2), (2, −3, 2), and (3, −4, 2).
How do you know if a linear system has many solutions?
Response: The system of an equation has infinitely many solutions when the lines are coincident, and they have the same y-intercept. If the two lines have the same y-intercept and the slope, they are actually in the same exact line.
How to tell how many solutions a linear system has without solving?
The answer is: If the slopes are the same but the y-intercepts are different, the system has no solution. If the slopes are different, the system has one solution. If the slopes are the same and the y-intercepts are the same, the system has infinitely many solutions.
Is a system of linear equations solvable?
As an answer to this: So "showing a system of linear equations isnot solvable" (has no solutions) is, by definition, the same thing as showing that the system of linear equations is "inconsistent".
How do you know if a linear system has exactly one solution?
If every variable is basic, then the linear system has exactly one solution. If two augmented matrices are row equivalent to one another, then they describe two linear systems having the same solution spaces. The presence of a free variable indicates that there are no solutions to the linear system.
What does it mean if a linear system has a free variable?
The presence of a free variable indicates that there are no solutions to the linear system. If a linear system has exactly one solution, then it must have the same number of equations as unknowns. If a linear system has the same number of equations as unknowns, then it has exactly one solution.
Does a consistent linear system have infinite solutions?
The response is: If a consistent linear system of equations has a free variable, it has infinite solutions. If a consistent linear system has more variables than leading 1s, then the system will have infinite solutions. A consistent linear system with more variables than equations will always have infinite solutions.
Is a system of linear equations solvable?
So "showing a system of linear equations isnot solvable" (has no solutions) is, by definition, the same thing as showing that the system of linear equations is "inconsistent".
How do you know if a system of linear equations is consistent?
Answer: A system of linear equations is consistent if it has a solution (perhaps more than one). A linear system is inconsistent if it does not have a solution. How can we tell what kind of solution (if one exists) a given system of linear equations has? The answer to this question lies with properly understanding the reduced row echelon form of a matrix.
What does it mean if a linear system has a free variable?
As an answer to this: The presence of a free variable indicates that there are no solutions to the linear system. If a linear system has exactly one solution, then it must have the same number of equations as unknowns. If a linear system has the same number of equations as unknowns, then it has exactly one solution.
How many solutions are there to a linear system?
As a response to this: Consider the following linear system: x − y = 0. There are obviously infinite solutions to this system; as long as x = y, we have a solution. We can picture all of these solutions by thinking of the graph of the equation y = x on the traditional x, y coordinate plane. Let’s continue this visual aspect of considering solutions to linear systems.

Rate article 