Respuesta :
Answer:
The system has infinitely many solutions.
[tex]\left\begin{array}{ccc}x_1&=&-\frac{1}{3}x_2-\frac{16}{9}x_4-\frac{2}{9} \\x_2&=&s_1\\x_3&=&-\frac{4}{3}x_4+\frac{1}{3} \\x_4&=&s_2\end{array}\right[/tex]
Step-by-step explanation:
To find the solution for this system of linear equations [tex]-3x_1-x_2+4x_3=2\\-3x_3 - 4x_4 = -1[/tex]you must:
Step 1: Transform the augmented matrix to the reduced row echelon form.
A matrix is a rectangular arrangement of numbers into rows and columns.
A system of equations can be represented by an augmented matrix.
In an augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.
This is matrix that represents the system
[tex]\left[ \begin{array}{ccccc} -3 & -1 & 4 & 0 & 2 \\\\ 0 & 0 & -3 & 4 & -1 \end{array} \right][/tex]
The augmented matrix can be transformed by a sequence of elementary row operations to the matrix.
There are three kinds of elementary matrix operations.
- Interchange two rows (or columns).
- Multiply each element in a row (or column) by a non-zero number.
- Multiply a row (or column) by a non-zero number and add the result to another row (or column).
Using elementary matrix operations, we get that
Row Operation 1: Multiply the 1st row by -1/3
Row Operation 2: Multiply the 2nd row by -1/3
Row Operation 3: Add 4/3 times the 2nd row to the 1st row
[tex]\left[ \begin{array}{ccccc} 1 & \frac{1}{3} & 0 & \frac{16}{9} & - \frac{2}{9} \\\\ 0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right][/tex]
Step 2: Interpret the reduced row echelon form
The reduced row echelon form of the augmented matrix is
[tex]\left[ \begin{array}{ccccc} 1 & \frac{1}{3} & 0 & \frac{16}{9} & - \frac{2}{9} \\\\ 0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right][/tex]
which corresponds to the system
[tex]x_1+\frac{1}{3}x_2+ \frac{16}{9}x_4=-\frac{2}{9} \\x_3+ \frac{4}{3}x_4=\frac{1}{3}[/tex]
We see that the variables [tex]x_2, x_4[/tex] can take arbitrary numbers; they are called free variables. Let [tex]x_2=s_1[/tex], [tex]x_4=s_2[/tex]. All solutions of the system are given by
[tex]\left\begin{array}{ccc}x_1&=&-\frac{1}{3}x_2-\frac{16}{9}x_4-\frac{2}{9} \\x_2&=&s_1\\x_3&=&-\frac{4}{3}x_4+\frac{1}{3} \\x_4&=&s_2\end{array}\right[/tex]
The system has infinitely many solutions.
The solutions of the system of equations are:
- x₁ = s₁
- x₂ = s₂
- x₃ = (2 + 3s₁ + s₂)/4
- x₄ = ((2 + 3s₁ + s₂)*(-3/4) + 1)/4
How to solve the system of equations?
Here we have the system:
-3x₁ - x₂ - 4x₃ = 2
-3x₃ - 4x₄ = -1
You can see that the only common variable between the two equations is x₃, so we can isolate that one, if we do it in the first equation we get:
x₃ = (-1 + 4x₄)/-3
Replacing that in the other equation we get:
-3x₁ - x₂ - 4*(-1 + 4x₄)/-3 = 2
Now, x₁ and x₂ are the free variables, so I will write them as:
s₁ and s₂, so we et:
-3s₁ - s₂ + 4*(-1 + 4x₄)/-3 = 2
Solving this for x₄ we get:
x₄ = ((2 + 3s₁ + s₂)*(-3/4) + 1)/4
Replacing that in the equation for x₃ we get:
x₃ = (-1 + 4x₄)/-3 = (2 + 3s₁ + s₂)/4
If you want to learn more about systems of equations, you can read:
https://brainly.com/question/13729904
