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Given vector a=(3,0), b=(K,5), and the angle between vector a and vector b is 3π/4, then the value of k is
Cos=(a*b)/(|a*|b|)=3k/3*Root (k^2+25)=-Root 2/2
K/Root (k^2+25)=-Root 2/2k
Determine the constants a so that the vector groups α1,α2 and α3 can be linearly represented by the vector groups β1,β2 and β3, but β1,β2 and β3 can not be linearly represented by α1,α2 and α3. Let α1=(1,1, a) T,α2=() T,α2=(1,a,1) T,α3=(a,1,1) T can be linearly represented by the vector group β1=(1,1, a) T,β2=(-2, a,4) T,β3=(-2, a, a) T, but the vector group β1,β2,β3 can not be linearly represented by the vector group α1,α2,α3.
If the rank of α is less than β, it is 3, so a can only be 1, or 2
When rank=1, a=1, it is obvious that the condition is met.
Solution 1 1A
1A1=0, this equation, besides 1, also has a root of -2, bring in β and find that β is not full rank, so discard
A1 1
Therefore a can only be equal to 1
If β=(0, k, k^2) can be expressed by α1=(1+k,1,1),α2=(1,1+k,1),α3=(1,1,1+k) linearly, then the value of k is -------? The answer is: k =0 and k =-3 How did you get here? Thank you.
β=(0, K, k^2) can be expressed by α1=(1+k,1,1),α2=(1,1+k,1),α3=(1,1,1+k) only linearly
So a system of linear equations x1α1+x2α2+x3α3=β has a unique solution
So |α1,α2,α3|=0
So (k+3) k^2=0
I.e. k =0 and k =-3
It is proved that if vector group α1 2 3 4,α5 is linearly independent, then vector group α1 2,α2 3,α3 4,α4 5,α5 1 is linearly independent? There's one more question. It is proved that if vector group α1 2 3 4 is linearly independent, then vector group α1 2,α2 3,α3 4,α4 1 is linearly dependent.
Let k1(α1 2)+k2(α2 3)+k3(α3 4)+k4(α4 5)+k5(α5 1)=0 k1,k2,k3,k4,k5 then (k1+k5)α1+(k1+k2)α2+(k1+k2)α3+(k2+k3)α3+(k3+k4)α4+(k4+k5)α5=0.
The vector α1α2α3 linearly independent α4 can not be expressed linearly by α1α2α3. How to prove that α1α2α3α4 linearly independent
Counter evidence
If α1α2α3α4 linear correlation
Because α1α2α3 is linearly independent
Thus α4 can be represented linearly by α1α2α3
Conflicts with known
So α1α2α3α4 is linearly independent
Given vector a=(1,0,2λ), vector b=(6,2μ-1,2)! Given vector a=(1,0,2λ), vector b=(6,2μ-1,2), if a//b, then the values of λ and μ are (), respectively The logical conjunction used in the proposition that the solution of equation x^2-2=0 is x=+-root number 2 is () Is the logical conjunction "OR" or "AND" or "NOT" used? ?? ?
A//b,(λ+1)/6=2λ/2,0=2μ-1
λ = 0.2 ,μ =0.5
Or
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