Question

Determine if this is a linearly dependent or independent set: (V, V, V ) where syon V = (1,2,2,-1), V = (4,9,9,-4), v = (5,8,9,-5). Show all work and Explain your reasoning carefully

Answer 1

Answer:

This is a linearly independent set.

Step-by-step explanation:

We have these following vectors:

$V_{1} = (1,2,2,-1)$

$V_{2} = (4,9,9,-4)$

$V_{3} = (5,8,9,-5)$

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:

$xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)$

If the only solution is $(x,y,z) = (0,0,0)$, they are L.I. Otherwise, they are L.D.

Solution:

$xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)$

$x(1,2,2,-1) + y(4,9,9,-4) + z (5,8,9,-5) = (0,0,0,0)$

We have the following system of equations:

$x + 4y + 5z = 0$

$2x + 9y + 8z = 0$

$2x + 9y + 9z = 0$

$-x -4y - 5z = 0$

I am going to solve this by the row-reduction of the augmented matrix.

This system has the following augmented matrix:

$\left[\begin{array}{cccc}1&4&5&0\\2&9&8&0\\2&9&9&0\\-1&-4&-5&0\end{array}\right]$

To reduce the first row, i am going to make these following operations:

$L_{2} = L_{2} - 2L_{1}$

$L_{3} = L_{3} - 3L_{1}$

$L_{4} = L_{4} + L_{1}$

So the augmented matrix now is:

$\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&1&-1&0\\0&0&0&0\end{array}\right]$

Now I reduce the second row, doing:

$L_{3} = L_{3} - L_{2}$

So the matrix is:

$\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&0&1&0\\0&0&0&0\end{array}\right]$

Now we can solve the system:

From the third line, we have that

$z = 0$

From the second line:

$y - 2z = 0$

$y - 2(0) = 0$

$y = 0$

From the first line

$x + 4y + 5z = 0$

$x + 4(0) + 5(0) = 0$

$x = 0$

The only solution for this system is $(x,y,z) = (0,0,0)$. This means that we have a linearly independent set.