N vector, X1, X2, X3, ...., Xn are linearly dependent if there exist scalars c1, c2, ...., cn, not all zero such that

$\displaystyle \sum^n_{i=1} c_i X_i = 0$

The given vector are

$X_1= [1, 2, -1, 0]$

$X_2 = [11, 0, 1, 2]$

$X_3=[4, 2, 1, 0]$ and

$X_4 = [6, 1, 0, 1]$

Assume that the given vectors are linearly dependent. Then,

$c_1 \times X_1 + c_2 \times X_2 + c_3 \times X_3 + c_4 \times X_4 = 0$

$\therefore c_1 \times X_1 + c_2 \times X_2 + c_3 \times X_3 = -c_4 \times X_4$

$\therefore \dfrac {-c_1}{c_4} \times X_1 + \dfrac {-c_2}{c_4} \times X_2 + \dfrac {-c_3}{c_4} \times X_3 = X_4$

Let $\dfrac {-c_1}{c_4} = a; \ \dfrac {-c_2}{c_4} =b ; \ \dfrac {-c_3}{c_4} = c$

$\therefore a \times X_1 + b\times X_2 + c\times X_3 = X_4$

$\therefore a\times [1, 2, -1, 0] + b\times [11, 0, 1, 2] + c \times [4, 2, 1, 0] = [6, 1, 0, 1]$

$\therefore [1a, 2a, -1a, 0] + [11b, 0, 1b, 2b] + [4c, 2c, 1c, 0] = [6, 1, 0, 1]$

$\therefore [a+11b+4c, 2a+2c, -a+b+c, 2b] = [6, 1, 0, 1]$

$\therefore [a+11b + 4c, 2a+2c, -a+b+c, 2b] = [6, 1, 0, 1]$

Equating the terms we get,

$1. \ \ a+11b+4c=6$

$2. \ \ 2a +2c = 1$

$3. \ \ -a+b+c=0$

$4. \ 2b=1$

On solving the equations (2), (3) and (4) we get, $a = \dfrac {1}{2}; \ b=\dfrac {1}{2} $ and $c=0.$

Substituting these values in equation (1).

$a+11b+4c = \dfrac {1}{2} + 11 \times \dfrac {1}{2}+ 4 \times 0 = 6 \cdots$satisfies

So the given vectors must be linearly dependent and their relation can be give

$\dfrac {1}{2}X_1 + \dfrac {1}{2} X_2 + 0 X_3 = X_4$