Jonathan's question at Yahoo Answers ( span ( H ) )

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In summary, the value of "a" for which vector v= [2,-3,4,a] is in the set H=span {[2,1,-3,2],[0,2,-2,-2],[0,0,3,3]} is a=9.
  • #1
Fernando Revilla
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Here is the question:

Find the value of "a" for which

v= [2,−3,4,a]

is in the set

H=span {[2,1,−3,2],[0,2,−2,−2],[0,0,3,3]} a = ?

Here is a link to the question:

Find the value of "a" for which? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
Hello jonathan,

Clearly
$$\mbox{rank }\begin{bmatrix}{2}&{-1}&{\;\:3}& \;\;2\\{0}&{\;\;2}&{-2}&-2\\{0}&{\;\;0}&{\;\;3}&\;\;3\end{bmatrix}=3$$

Then, $(2,-3,4,a)$ belongs to $H$ if and only if:

$$\mbox{rank }\begin{bmatrix}{2}&{-3}&{\;\:4}& \;\;a\\{2}&{-1}&{\;\:3}& \;\;2\\{0}&{\;\;2}&{-2}&-2\\{0}&{\;\;0}&{\;\;3}&\;\;3\end{bmatrix}=3$$
Equivalently, if and only if
$$\det\;\begin{bmatrix}{2}&{-3}&{\;\:4}& \;\;a\\{2}&{-1}&{\;\:3}& \;\;2\\{0}&{\;\;2}&{-2}&-2\\{0}&{\;\;0}&{\;\;3}&\;\;3\end{bmatrix}=0$$

Easily we verify that the previous determinant is $12(9-a)$ so,

$(2,-3,4,a)$ belongs to $H$ if and only if $a=9$
 

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