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Math  /  Algebra

Question8. Consider the following homogeneous system of linear equations in four variables. For convenience, the general solution is given. Show that the set of solutions forms a subspace of R4\mathbf{R}^{4}. x1+x2−3x3+5x4=0x2−x3+3x4=0x1+2x2−4x3+8x4=0\begin{aligned} x_{1}+x_{2}-3 x_{3}+5 x_{4} & =0 \\ x_{2}-x_{3}+3 x_{4} & =0 \\ x_{1}+2 x_{2}-4 x_{3}+8 x_{4} & =0 \end{aligned}
General solution is (2r−2s,r−3s,r,s)(2 r-2 s, r-3 s, r, s).

Studdy Solution

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)(2r - 2s, r - 3s, r, s).
3. We need to show that the set of solutions forms a subspace of R4\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=0r = 0 and s=0s = 0 into the general solution:
(2(0)−2(0),0−3(0),0,0)=(0,0,0,0) (2(0) - 2(0), 0 - 3(0), 0, 0) = (0, 0, 0, 0)
The zero vector (0,0,0,0)(0, 0, 0, 0) is indeed a solution.

STEP 4

Check closure under addition.
Consider two arbitrary solutions (2r1−2s1,r1−3s1,r1,s1)(2r_1 - 2s_1, r_1 - 3s_1, r_1, s_1) and (2r2−2s2,r2−3s2,r2,s2)(2r_2 - 2s_2, r_2 - 3s_2, r_2, s_2).
Add these two solutions:
(2r1−2s1,r1−3s1,r1,s1)+(2r2−2s2,r2−3s2,r2,s2)=(2(r1+r2)−2(s1+s2),(r1+r2)−3(s1+s2),r1+r2,s1+s2)\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)(2r - 2s, r - 3s, r, s), where r=r1+r2r = r_1 + r_2 and s=s1+s2s = s_1 + s_2, which is a solution.

STEP 5

Check closure under scalar multiplication.
Consider a solution (2r−2s,r−3s,r,s)(2r - 2s, r - 3s, r, s) and a scalar cc.
Multiply the solution by the scalar:
c⋅(2r−2s,r−3s,r,s)=(c(2r−2s),c(r−3s),cr,cs)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′)(2r' - 2s', r' - 3s', r', s'), where r′=crr' = cr and s′=css' = 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 R4\mathbf{R}^{4}.

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