Perform some computations on vectors in R4

Given vectors

$A = (1,1,1,0), \qquad B = (0,1,1,1), \qquad C = (1,1,0,0)$

in $\mathbb{R}^4$ define $D= xA + yB + zC$ for scalars $x,y,z \in \mathbb{R}$ .

The components of $D$ are
$D = x(1,1,1,0) + y(0,1,1,1) + z(1,1,0,0) = (x+z, x+y+z, x+y, y).$
Proof. If $D = O = (0,0,0,0)$ then we have
$\begin{align*} x+z &=0 \\ x+y+z &= 0 \\ x+y &= 0 \\ y &= 0. \end{align*}$

From the third equation and $y =0$ we have $x =0$ , and then the first equation implies $z = 0$ . Hence, we must have $x = y = 0. \qquad \blacksquare$
If $D = (1,5,3,4)$ then we have
$\begin{align*} x + z &= 1 \\ x+y+z &= 5 \\ x+y &=3 \\ y &= 4. \end{align*}$

Since $y = 4$ and $x+y = 3$ we must have $x = -1$ . Then since $x+z = 1$ we have $z = 2$ . Therefore, $x = -1, y = 4, z = 2$ are values such that $D = (1,5,3,4)$ .
If $D = (1,2,3,4)$ then we must have
$\begin{align*} x+z &= 1 \\ x+y+z &= 2 \\ x+y &= 3 \\ y &= 4. \end{align*}$

But, $y= 4$ and $x + y = 3$ implies $x = -1$ . Then, $x+z = 1$ would implies $z = 2$ . However, then we run into a contradiction since $x+y+z = -1 + 4 + 2 = 5 \neq 2. \qquad \blacksquare$