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There is a column of numbers: A1, A2, A3 If A1 = 2, then a2007 is () A. 2007B. 2C. 12D. -1
According to the meaning of the question: A1 = 2, A2 = 1-12 = 12, A3 = 1-2 = - 1, A4 = 1 + 1 = 2; period is 3; 2007 △ 3 = 669; so a2007 = A3 = - 1
If A1 = - 1 / 2, starting from the second number, each number is equal to the reciprocal of the difference between 1 and the number in front of it, find the values of A2, A3, A4, A20, A2010
a2=1/(1-(-1/2))=2/3
a3=1/(1-2/3))=3
a4=1/(1-3)=-1/2
So this sequence is a cyclic sequence
a1=a4=a7...
a2=a5=a8...
a3=a6=a9...
therefore
a20=a2=2/3
A2010 = A3 (because it can be divided by 3) = 3
Let n-dimensional vector group A1, A2,..., am be linearly independent, and A1, A2,..., am, B be linearly related,
Let n-dimensional vector group A1, A2,..., am be linearly independent, and A1, A2,..., am, B be linearly related. Two different methods are used to prove that B can be linearly represented by A1, A2,..., am, and the representation is unique
Proof 1. Because A1, A2,..., am and B are linearly related, there exists a group of numbers K1, K2,..., KM which are not all zero. If K makes k1a1 + k2a2 +... + kmam + KB = 0, then K ≠ 0 must exist. Otherwise, k1a1 + k2a2 +... + kmam = 0, and A1, A2,..., am are linearly independent, so K1 = K2 =... = km = 0
Let a be a positive definite matrix of order n, A1, A2. Am be n-dimensional non-zero column vector, and AI ^ taaj = 0. It is proved that A1, A2. Am are linearly independent
The definition of positive definite
If x! = 0, then x'ax > 0
The title is wrong
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