Question

Solve the following sets of simultaneous equations by reducing the matrix to row echelon form. $$ \left\\{\begin{aligned} x-3 y-2 z+& 3 v &=1 \\ z+3 u+5 v &=-1 \\ x-3 y+z+u+2 v &=6 \\ z+u+v &=1 \end{aligned}\right. $$

Short answer

Row echelon form achieved. Solutions can be found with back-substitution.

Step-by-step solution

Write the augmented matrix

Start by writing the system of linear equations as an augmented matrix. The given equations are: \[ \begin{aligned} x - 3y - 2z + 3v &= 1 \ z + 3u + 5v &= -1 \ x - 3y + z + u + 2v &= 6 \ z + u + v &= 1 \end{aligned} \] The augmented matrix is: \[ \begin{bmatrix} 1 & -3 & -2 & 0 & 3 & | & 1 \ 0 & 0 & 1 & 3 & 5 & | & -1 \ 1 & -3 & 1 & 1 & 2 & | & 6 \ 0 & 0 & 1 & 1 & 1 & | & 1 \end{bmatrix} \]

Simplify the first two rows

Subtract the first row from the third row to eliminate the x-term in the third row: \[ \begin{bmatrix} 1 & -3 & -2 & 0 & 3 & | & 1 \ 0 & 0 & 1 & 3 & 5 & | & -1 \ 0 & 0 & 3 & 1 & -1 & | & 5 \ 0 & 0 & 1 & 1 & 1 & | & 1 \end{bmatrix} \]

Further reduce the matrix

Subtract the third row from the fourth row to eliminate the z-term in the fourth row: \[ \begin{bmatrix} 1 & -3 & -2 & 0 & 3 & | & 1 \ 0 & 0 & 1 & 3 & 5 & | & -1 \ 0 & 0 & 3 & 1 & -1 & | & 5 \ 0 & 0 & 0 & 2 & 2 & | & -4 \end{bmatrix} \]

Normalize the last row

Divide the last row by 2 to facilitate further operations: \[ \begin{bmatrix} 1 & -3 & -2 & 0 & 3 & | & 1 \ 0 & 0 & 1 & 3 & 5 & | & -1 \ 0 & 0 & 3 & 1 & -1 & | & 5 \ 0 & 0 & 0 & 1 & 1 & | & -2 \end{bmatrix} \]

Eliminate other terms

Use the last row to eliminate the u-term in the second and third rows: Subtract 3 times the fourth row from the second row and 1 times the fourth row from the third row to get: \[ \begin{bmatrix} 1 & -3 & -2 & 0 & 3 & | & 1 \ 0 & 0 & 1 & 0 & 2 & | & 5 \ 0 & 0 & 3 & 0 & -2 & | & 7 \ 0 & 0 & 0 & 1 & 1 & | & -2 \end{bmatrix} \]

Normalize second and third rows

Divide the third row by 3 and the second row by 1: \[ \begin{bmatrix} 1 & -3 & -2 & 0 & 3 & | & 1 \ 0 & 0 & 1 & 0 & 2 & | & 5 \ 0 & 0 & 1 & 0 & -2/3 & | & 7/3 \ 0 & 0 & 0 & 1 & 1 & | & -2 \end{bmatrix} \]

Reduce to row echelon form

Subtract the second row from the first row to eliminate the z-term in the first row: \[ \begin{bmatrix} 1 & -3 & 0 & 0 & 5 & | & -4 \ 0 & 0 & 1 & 0 & 2 & | & 5 \ 0 & 0 & 1 & 0 & -2/3 & | & 7/3 \ 0 & 0 & 0 & 1 & 1 & | & -2 \end{bmatrix} \] Now, using the second (z-eliminated) row, eliminate the z-values in the third row: \[ \begin{bmatrix} 1 & -3 & 0 & 0 & 5 & | & -4 \ 0 & 0 & 1 & 0 & 2 & | & 5 \ 0 & 0 & 0 & 0 & -8/3 & | & -4/3 \ 0 & 0 & 0 & 1 & 1 & | & -2 \end{bmatrix} \]

Key concepts

matrix, simultaneous equations

The set of simultaneous equations consists of several equations that are solved together. For instance, the provided equations involving variables x, y, z, u, and v are simultaneous because their solutions must satisfy all the equations at once. Matrices provide a powerful tool to solve these equations efficiently.
In this context, we express the system of equations as a matrix. Each equation corresponds to a row, and each variable corresponds to a column of the matrix. The matrix representation simplifies operations like addition, subtraction, and scaling, making it easier to manipulate and solve the equations simultaneously.

augmented matrix

An augmented matrix is a combination of the coefficient matrix of a system of linear equations and the constants from the equations. By incorporating the constants as an additional column, we get a complete representation of the system.
In the given problem, the original equations are transformed into the following augmented matrix:
```
[ 1 -3 -2 0 3 | 1 ]
[ 0 0 1 3 5 | -1 ]
[ 1 -3 1 1 2 | 6 ]
[ 0 0 1 1 1 | 1 ]
```
This matrix form allows us to use row operations to systematically reduce the system to row echelon form, simplifying the process of finding the solution.

linear algebra

Linear algebra is the branch of mathematics that deals with vector spaces and linear mappings between them. It is fundamental in solving systems of linear equations, which are essential in various fields of science and engineering.
The main goal in the given problem is to transform the augmented matrix into row echelon form using linear algebra techniques. This involves:

• Eliminating variables step-by-step through row operations
• Normalizing rows for easier manipulation
• Achieving a form where back-substitution can be easily applied

Row echelon form is a simplified version of the matrix where the solution to the system is clearer and can be derived directly, thus demonstrating the power and utility of linear algebra.