find the system of linear equations

Question

Find the system of linear equations, that has subspace of solutions similar to linear span of vectors system $a_1, a_2, a_3$, where:
$a_1 = (2, 1, -1, 1)$,
$a_2 = (1, 2, 1, -1)$,
$a_3 = (1, -4, -5, 5)$

Answer 1

$$ x - y + z = 0 \\ -x + y + w = 0 $$

How'd I do that? I said that I needed vectors which, when multiplied by the three vectors you wrote, gave zeros. So I wrote down the matrix $M$

2   1  -1   1
1   2   1  -1
1  -4  -5   5

and said that I wanted vectors $b$ with $Mb = 0$. These vectors, when transposed, would be the coefficients of the equations I needed. To find these, I did row operations on the vectors, swapping the first two:

1   2   1  -1
2   1  -1   1
1  -4  -5   5

and then subtracting twice the first from the second (this doesn't change the solution vectors $b$) to get

1   2   1  -1
0  -3  -3   3
1  -4  -5   5

and then subtracting the first from the third:

1   2   1  -1
0  -3  -3   3
0  -6  -6   6

Subtracting twice the second from the third gives a row of zeros, which I tossed out. So I got a $2 \times 4$ matrix $Q$,

1   2   1  -1
0  -3  -3   3

and want to solve $Q\begin{bmatrix}a\\b\\c\\d\end{bmatrix} = 0$. To do so, I first set $c = 1$ and $ d = 0$, and solved for $a$ and $b$; then I set $c = 0, d = 1$, and solved for $a$ and $b$. The results were, respectively $$ (a, b, c, d) = (1, -1, 1, 0) \\ (a, b, c, d) = (-1, 1, 0, 1) $$ And that led to the two equations I wrote down.