Question

63. Show that a necessary and sufficient condition that the vectors A = 4,1 + A21+ A, k, B = B, i+B,1+Bok,
A2 A2 A3
C = C i +C2j+Cgk be linearly independent is that the determinant B, B2 B3 be different from zero.
C, C2
C3
64. (e) Prove that the yestors
A- Pil​

Answer 1

Answer:

Start with assuming that λ1(1,a,a2)+λ2(1,b,b2)+λ3(1,c,c2)=0. Then this leads to a system of homogeneous linear equations with coefficient matrix ⎛⎝⎜1aa21bb21cc2⎞⎠⎟.

And the determinant of this matrix is (b−a)(c−b)(c−a) which is nonzero. Hence the system only have zero solution.

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answered

Mar 5 '15 at 15:46

Tim

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The stated vectors form a Vandermonde Matrix, whose determinant is zero (i.e. linearly dependent rows/columns) if, and only if, either a=b, a=c or b=c. .

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Mar 5 '15 at 15:43

ki3i

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If you don't know where to start you should ask yourself what does it mean for some vectors to be linearly independent? A quick Google search will give you definitions and worked out examples.

Given the vectors x=⎡⎣⎢1aa2⎤⎦⎥, y=⎡⎣⎢1bb2⎤⎦⎥, z=⎡⎣⎢1cc2⎤⎦⎥, you need to show that, given a≠b≠c and r,s,t∈R, the equation

rx+sy+tz=0

can only be satisfied when r=s=t=0. That gives you the following three questions:

r+s+t=0, ra+sb+tc=0, and ra2+sb2+tc2=0. Is it possible to solve all three equations with some nonzero r,s, or t? You need to show that you CAN'T find a non-zero r,s or t. You could also show it pairwise: first show x,y are linearly independent, then x,z, and then y,z.