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.

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…

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**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.

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