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Let

Find a basis of a space

I think the question is quite easy. Given this vector

I believe there is something wrong. Things above are what I can compute...

Is there any other way to solve this question?

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Oh, I've thought of another way.

Set a augmented matrix

[1 2 3 -1 2 | 0]

[2 4 7 2 -1 | 0]

~

[1 2 3 -1 2 | 0]

[0 2 4 3 -3 | 0]

Let x

∴ x

x

Therefore, the solution set is {t[-5 1.5 2 0 0 1]

The basis is {[-5 1.5 2 0 0 1]

Is this correct?

**u**= [1, 2, 3, -1, 2]^{T},**v**= [2, 4, 7, 2, -1]^{T}in**ℝ**^{5}**.**Find a basis of a space

*W*such that**w**⊥**u**and**w**⊥**v**for all**w**∈*W.*I think the question is quite easy. Given this vector

**w**in the space W is orthogonal to both**u**and**v**. I can only think of**w**being a__zero vector__. But would this be too trivial?**=****w**^{T}u**= 0****w**^{T}v**w**(^{T}**u**-**v**) = 0I believe there is something wrong. Things above are what I can compute...

Is there any other way to solve this question?

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Oh, I've thought of another way.

Set a augmented matrix

[1 2 3 -1 2 | 0]

[2 4 7 2 -1 | 0]

~

[1 2 3 -1 2 | 0]

[0 2 4 3 -3 | 0]

Let x

_{5}= t, x_{4}= s, x_{3}= z where t, s, z ∈ ℝ.∴ x

_{2}= 1.5t - 1.5s - 2zx

_{1}= -5t + 4s + zTherefore, the solution set is {t[-5 1.5 2 0 0 1]

^{T}+ s[4 -1.5 0 1 0]^{T}+ z[1 -2 1 0 0]^{T}}.The basis is {[-5 1.5 2 0 0 1]

^{T}, [4 -1.5 0 1 0]^{T}, [1 -2 1 0 0]^{T}}.Is this correct?

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