>>666
転置は関係ないので t は無視。
掃き出しの要領で
Ｗ1＝＜(2,1,1,0),(2,−1,−3,2)＞＝{x(2,1,1,0)＋y(2,−1,−3,2)|x,y∈R}＝{x(2,1,1,0)＋y(0,−2,−4,2)|x,y∈R}
　＝{x(2,1,1,0)＋y(0,1,2,−1)|x,y∈R}＝{x(2,0,−1,1)＋y(0,1,2,−1)|x,y∈R}
Ｗ2＝＜(2,1,−2,3),(1,1,0,1)＞＝{z(2,1,−2,3)＋w(1,1,0,1)|z,w∈R}＝{z(2,1,−2,3)＋w(2,2,0,2)|z,w∈R}
　＝{z(2,1,−2,3)＋w(0,1,2,−1)|z,w∈R}＝{z(2,0,−4,4)＋w(0,1,2,−1)|z,w∈R}＝{z(1,0,−2,2)＋w(0,1,2,−1)|z,w∈R}

Ｗ1∩Ｗ2＝{x(2,0,−1,1)＋y(0,1,2,−1)|x,y∈R,∃z,w∈R[x(2,0,−1,1)＋y(0,1,2,−1)＝z(1,0,−2,2)＋w(0,1,2,−1)]}
　　　　＝{x(2,0,−1,1)＋y(0,1,2,−1)|x,y∈R,∃w∈R[x(2,0,−1,1)＋y(0,1,2,−1)＝2x(1,0,−2,2)＋w(0,1,2,−1)]}
　　　　＝{x(2,0,−1,1)＋y(0,1,2,−1)|x,y∈R,∃w∈R[x(0,0,3,−3)＋y(0,1,2,−1)＝w(0,1,2,−1)]}
　　　　＝{x(2,0,−1,1)＋y(0,1,2,−1)|x,y∈R,x(0,0,3,−3)＝0}＝{y(0,1,2,−1)|y∈R}＝＜(0,1,2,−1)＞

Ｗ1＋Ｗ2＝{x(2,0,−1,1)＋y(0,1,2,−1)＋z(1,0,−2,2)＋w(0,1,2,−1)|x,y,z,w∈R}
　　　　＝{x(2,0,−1,1)＋z(1,0,−2,2)＋y(0,1,2,−1)|x,y,z∈R}＝{x(2,0,−1,1)＋z(2,0,−4,4)＋y(0,1,2,−1)|x,y,z∈R}
　　　　＝{x(2,0,−1,1)＋z(0,0,−3,3)＋y(0,1,2,−1)|x,y,z∈R}＝{x(2,0,−1,1)＋z(0,0,−1,1)＋y(0,1,2,−1)|x,y,z∈R}
　　　　＝{x(2,0,0,0)＋z(0,0,−2,2)＋y(0,1,2,−1)|x,y,z∈R}＝{x(1,0,0,0)＋z(0,0,−2,2)＋y(0,1,0,1)|x,y,z∈R}
　　　　＝{x(1,0,0,0)＋y(0,1,0,1)＋z(0,0,1,−1)|x,y,z∈R}＝＜(1,0,0,0),(0,1,0,1),(0,0,1,−1)＞