Verifying whether my working is correct in showing Linear Independence

[savva]

Homework Statement

I have attempted the questions below but am not sure if I am applying the method correctly to show linear dependence/independence.

a)Show that the vectors
e1=[1 1 0]T, e2=[1 0 1]T, e3=[0 1 1]T
are linearly independent

b) Show that the vectors
e1=[1 1 0]T, e2=[1 0 -1]T, e3=[0 1 1]T
are linearly independent

(T = Transverse)

Homework Equations

The determinant

The Attempt at a Solution

I have attempted to find the determinant by putting the vectors in a 3x3 matrix and finding the determinant which when =0 should give linear dependence and when ≠0 give linear independence. My working and the questions are attached in a pdf file with this thread.

I'd greatly appreciate any help

 

[HallsofIvy]

Do you know the basic definition of "independent", "dependent" vectors?

A set of vectors \{v_1, v_2, \cdot\cdot\cdot, v_n\} is "independent" if and only if in order to have a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n\}, we must have a_1= a_2= \cdot\cdot\cdot+ a_n= 0.

Here, such a sum would be of the form
a_1\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}+ a_2\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}+ a_3\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}0\\ 0 \\ 0 \end{bmatrix}
Of course multiplying the scalars and adding that is the same as
\begin{bmatrix}a_1+ a_2 \\ a_1+ a_3 \\a_1+ a_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}

which, in turn, is equivalent to the three equations
a_1+ a_2= 0, a_1+ a_3= 0, a_1+ a_2= 0

a_1= a_2= a_3= 0 is obviously a solution to that system of equations. Is it the only one (if so the vectors are independent. If there exist another, non "trivial" solution, they are dependent).

Of course, one can determine whether or not a system of equations is has unique solution by looking at the determinant of coefficients. As you say, these sets of vectors are independent (I would not say "independence exists").

 

[savva]

Do you know the basic definition of "independent", "dependent" vectors?

A set of vectors \{v_1, v_2, \cdot\cdot\cdot, v_n\} is "independent" if and only if in order to have a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n\}, we must have a_1= a_2= \cdot\cdot\cdot+ a_n= 0.

Here, such a sum would be of the form
a_1\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}+ a_2\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}+ a_3\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}0\\ 0 \\ 0 \end{bmatrix}
Of course multiplying the scalars and adding that is the same as
\begin{bmatrix}a_1+ a_2 \\ a_1+ a_3 \\a_1+ a_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}

which, in turn, is equivalent to the three equations
a_1+ a_2= 0, a_1+ a_3= 0, a_1+ a_2= 0

a_1= a_2= a_3= 0 is obviously a solution to that system of equations. Is it the only one (if so the vectors are independent. If there exist another, non "trivial" solution, they are dependent).

Of course, one can determine whether or not a system of equations is has unique solution by looking at the determinant of coefficients. As you say, these sets of vectors are independent (I would not say "independence exists").

I do not understand by what you mean by "As you say, these sets of vectors are independent (I would not say "independence exists")"
From my calculations I found the first question to be independent and the second dependent. What do you mean you would not say independence exists?

I don't understand how a1+a2=0 or a1+a3=0
if you add these up;
a1+a2= [1 1 0] + [1 0 1] = [2 1 1]
a1+a3= [1 1 0] + [0 1 1] = [1 2 1]

how do you get them to = 0?