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Question

For the given problem, consider the homogeneous systems of linear equations. For convenience, general solutions are given. It can be shown, as in the previous exercise, that the sets of solutions form subspaces of R4\mathrm{R}^4.

(a) Use the general solution to construct two specific solutions. (b) Use the operations of addition and scalar multiplication to generate four vectors from these two solutions. (c) Use the general solution to check that these vectors are indeed solutions, giving the values of rr and ss for which they are solutions.

x1+x2+x3+3x4=0x1+2x2+4x3+5x4=0x12x3+x4=0\begin{aligned} x_1+x_2+x_3+3 x_4&=0\\ x_1+2 x_2+4 x_3+5 x_4&=0\\ x_1 \quad-2 x_3+x_4&=0 \end{aligned}

General solution is (2rs,3r2s,r,s)(2 r-s,-3 r-2 s, r, s).

Solution

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For the system

x1  +  x2  +  x3  +  3x4=0x1  +  2x2  +  4x3  +  5x4=03x1  2x3  +  x4=0\begin{alignedat}{4} x_1 \; &+& \; x_2 \; &+& \; x_3 \; &+&\; 3x_4&=0\\ x_1 \; &+& \; 2x_2 \; &+& \; 4x_3 \; &+&\; 5x_4&=0\\ 3x_1 && &-& \; 2x_3 \; &+&\; x_4&=0\\ \end{alignedat}

we are given the general solution (2rs,3r2s,r,s)(2r-s,-3r-2s,r,s).

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