Let vector group A1, A2, A3 As is linearly independent (s > 2), try to prove that the following vector groups are vector independent: A1, a1 + A2, a1 + A2 + a3 a1+a2+…… as

Let vector group A1, A2, A3 As is linearly independent (s > 2), try to prove that the following vector groups are vector independent: A1, a1 + A2, a1 + A2 + a3 a1+a2+…… as

Let k1a1 + K2 (a1 + A2) + K3 (a1 + A2 + a3) +... + KS (a1 + A2 +... + as) = 0, then (K1 + K2 +... + KS) a1 + (K2 + K3 +... + KS) A2 +... + KSAs = 0. Given that A1, A2, A3,..., as are linearly independent, so K1 + K2 +... + KS = 0k2 +... + KS = 0... KS = 0, the solution K1 = K2 = K3 =... = KS = 0, so A1, a1 + A2

If the vector group A: A1, A2, A3, A4, A5 are linearly independent, the rank Ra of the vector group is zero=

If the vector group A: A1, A2, A3, A4, A5 are linearly independent, the rank of the vector group RA = 5

A1.a2.a3.a4 are the four different points given on the plane, so that the vector MA1 + ma2 + Ma3 + ma4 = 0 holds. What is the number of points m?

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Let A1, A2, A3 and A4 be given four different points on the plane, then the number of points m with vector MA1 + vector ma2 + vector Ma3 + vector ma4 = 0 is?

Let AI (Xi, Yi) (I = 1,2,3,4), m (x, y), then (oa1-om) + (oa2-om) + (oa3-om) + (oa4-om) = 0, and OM = (OA1 + oa2 + oa3 + oa4) / 4, that is, x = (x1 + x2 + X3 + x4) / 4, y = (Y1 + Y2 + Y3 + Y4) / 4. Therefore, there is exactly one point m satisfying the condition