Math Snap

STEP 1

1. We are given a homogeneous system of linear equations.
2. The general solution is provided as $(2r - 2s, r - 3s, r, s)$.
3. We need to show that the set of solutions forms a subspace of $\mathbf{R}^{4}$.

STEP 2

1. Verify that the zero vector is a solution.
2. Check closure under addition.
3. Check closure under scalar multiplication.

STEP 3

Verify that the zero vector is a solution.
Substitute $r = 0$ and $s = 0$ into the general solution:
$$(2(0) - 2(0), 0 - 3(0), 0, 0) = (0, 0, 0, 0)$$ The zero vector $(0, 0, 0, 0)$ is indeed a solution.

STEP 4

Check closure under addition.
Consider two arbitrary solutions $(2r_1 - 2s_1, r_1 - 3s_1, r_1, s_1)$ and $(2r_2 - 2s_2, r_2 - 3s_2, r_2, s_2)$.
Add these two solutions:
$$\begin{aligned} &(2r_1 - 2s_1, r_1 - 3s_1, r_1, s_1) + (2r_2 - 2s_2, r_2 - 3s_2, r_2, s_2) \\ &= (2(r_1 + r_2) - 2(s_1 + s_2), (r_1 + r_2) - 3(s_1 + s_2), r_1 + r_2, s_1 + s_2) \end{aligned}$$ This is of the form $(2r - 2s, r - 3s, r, s)$, where $r = r_1 + r_2$ and $s = s_1 + s_2$, which is a solution.

Check closure under scalar multiplication.
Consider a solution $(2r - 2s, r - 3s, r, s)$ and a scalar $c$.
Multiply the solution by the scalar:
$$c \cdot (2r - 2s, r - 3s, r, s) = (c(2r - 2s), c(r - 3s), cr, cs)$$ This is of the form $(2r' - 2s', r' - 3s', r', s')$, where $r' = cr$ and $s' = cs$, which is a solution.
Since the zero vector is a solution, and the set is closed under addition and scalar multiplication, the set of solutions forms a subspace of $\mathbf{R}^{4}$.