Respuesta :
Answer:
This is a linearly independent set.
Step-by-step explanation:
We have these following vectors:
[tex]V_{1} = (1,2,2,-1)[/tex]
[tex]V_{2} = (4,9,9,-4)[/tex]
[tex]V_{3} = (5,8,9,-5)[/tex]
In a set of 3 vectors, if one of these vectors can be written as a linear combination of the 2 other vectors, they are linearly dependent. Otherwise, they are linearly independent.
We can verify this by solving the following system:
[tex]xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)[/tex]
If the only solution is [tex](x,y,z) = (0,0,0)[/tex], they are L.I. Otherwise, they are L.D.
Solution:
[tex]xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)[/tex]
[tex]x(1,2,2,-1) + y(4,9,9,-4) + z (5,8,9,-5) = (0,0,0,0)[/tex]
We have the following system of equations:
[tex]x + 4y + 5z = 0[/tex]
[tex]2x + 9y + 8z = 0[/tex]
[tex]2x + 9y + 9z = 0[/tex]
[tex]-x -4y - 5z = 0[/tex]
I am going to solve this by the row-reduction of the augmented matrix.
This system has the following augmented matrix:
[tex]\left[\begin{array}{cccc}1&4&5&0\\2&9&8&0\\2&9&9&0\\-1&-4&-5&0\end{array}\right][/tex]
To reduce the first row, i am going to make these following operations:
[tex]L_{2} = L_{2} - 2L_{1}[/tex]
[tex]L_{3} = L_{3} - 3L_{1}[/tex]
[tex]L_{4} = L_{4} + L_{1}[/tex]
So the augmented matrix now is:
[tex]\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&1&-1&0\\0&0&0&0\end{array}\right][/tex]
Now I reduce the second row, doing:
[tex]L_{3} = L_{3} - L_{2}[/tex]
So the matrix is:
[tex]\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&0&1&0\\0&0&0&0\end{array}\right][/tex]
Now we can solve the system:
From the third line, we have that
[tex]z = 0[/tex]
From the second line:
[tex]y - 2z = 0[/tex]
[tex]y - 2(0) = 0[/tex]
[tex]y = 0[/tex]
From the first line
[tex]x + 4y + 5z = 0[/tex]
[tex]x + 4(0) + 5(0) = 0[/tex]
[tex]x = 0[/tex]
The only solution for this system is [tex](x,y,z) = (0,0,0)[/tex]. This means that we have a linearly independent set.
