Let S be all solution to x1 x2 x3 x4 0 x1 2x2 5x3 4x4

Let S be all solution to x_1 - x_2 - x_3 + x_4 = 0 x_1 + 2x_2 + 5x_3 + 4x_4 = 0 Find a basis for S and S Use G. S to make basis m both on b of S and S respectively State why the on b are then an on b for R^4.

Solution

2(a) Let x₃ = r and x₄ = t. Then x₁ –x₂ = r-t …(1) and x₁ +2x₂ = -5r -4t…(2). On subtracting th 1^st equation from the 2^nd equation, we get x₁ +2x₂ –(x₁ –x₂) = -5r -4t –(r-t) or, 3x₂ = -6r-3t so that x₂ = -2r-t. On substituting this value of x₂ in the 1^st equation, we get x₁ = r-t+( -2r-t) = -r . Then X = (x₁,x₂,x₃,x₄)^T =( -r, -2r-t, r,t)^T= r(-1, -2,1,0)^T + t( 0,-1,0,1)^T . Hence a basis for S is the set { (-1, -2,1,0)^T, ( 0,-1,0,1)^T}.

Let (x,y,z,w)^T be an arbitrary element of S. Then (x, y, z,w).(-1,-2,1,0) = 0 or, -x -2y +z = 0 and (x, y, z,w). ( 0,-1,0,1) = 0 or, -y +w = 0. Let z = p and w = q. Then y = q, -x-2q +p = 0 so that x = -2q+p. Then, (x,y,z,w)^T = ( p-2q ,q, p, q)^T = p(1,0,1,0)^T +q( -2,1,0,1)^T. Hence a basis for S is the set { (1,0,1,0)^T, ( -2,1,0,1)^T }.

(b) & (c). The meaning of on b is not clear. Please mention clearly in your comment so that these questions may be answered.