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+x23x3+5x4=0x2x3+3x4=0x1+2x24x3+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 (2r2s,r3s,r,s)(2 r-2 s, r-3 s, r, s).

Studdy Solution
Check closure under scalar multiplication.
Consider a solution (2r2s,r3s,r,s)(2r - 2s, r - 3s, r, s) and a scalar cc.
Multiply the solution by the scalar:
c(2r2s,r3s,r,s)=(c(2r2s),c(r3s),cr,cs)c \cdot (2r - 2s, r - 3s, r, s) = (c(2r - 2s), c(r - 3s), cr, cs)
This is of the form (2r2s,r3s,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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