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Given a procession A = eginbmatrix1 &2 &3 \ 4 &5 &6 \ 7 &8 &9 endbmatrix

Determine if vector $b$ is in $span(A)$where $$b = eginbmatrix1 \ 2 \ 4 endbmatrix$$


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First we deal with $ hedesigningfairy.comrmSpan(A)$,

$$ hedesigningfairy.comrmSpan(A) = left left< eginarrayc x_1\ x_2\ x_3\ endarray ight> : left< eginarrayc x_1\ x_2\ x_3\ endarray ight> = a left<eginarrayc 1\ 4\ 7\ endarray ight> + b left<eginarrayc 2\ 5\ 8\ endarray ight> + c left<eginarrayc 3\ 6\ 9\ endarray ight>. A,b,c in hedesigningfairy.combbR ight.$$

From this an interpretation we can see the asking if vector $vecb in hedesigningfairy.comrmSpan(A)$ is equivalent to questioning if over there exists a vector $vecx$ such the $Avecx = vecb$,

because $$vecx = left<eginarrayc x_1\ x_2\ x_3\ endarray ight>,$$

then $$Avecx = left<eginarrayccc 1 & 2 & 3\ 4& 5 & 6\ 7 & 8 & 9\ endarray ight> left< eginarrayc x_1\ x_2\ x_3 endarray ight> = x_1 left< eginarrayc 1\ 4\ 7 endarray ight> + x_2 left< eginarrayc 2\ 5\ 8 endarray ight> + x_3 left< eginarrayc 3\ 6\ 9 endarray ight>.$$

So if over there exists $x_1, x_2, x_3$ such the the final line amounts to $vecb$, climate we recognize $vecb$ is in $ hedesigningfairy.comrmSpan(A)$. This way it suffices to ask if over there exists a $vecx$ such that $Avecx = vecb$.

This last equation is identical to the procession equation:

$$left<eginarrayccc 1 & 2 & 3\ 4 & 5 & 6\ 7 & 8 & 9 endarray ight> vecx = left<eginarrayc 1\ 2\ 4\ endarray ight>,$$which we can convert to the system:

$$left<eginarrayc 1 & 2 & 3 & 1\ 4 & 5 & 6 & 2\ 7 & 8 & 9 & 4\endarray ight>.$$

As you can have learned, we solve this mechanism by heat reduction (I used an innovation for this step, your instructor may require row reduction through hand):

$$ hedesigningfairy.comrmRREFleft(left<eginarrayc 1 & 2 & 3 & 1\ 4 & 5 & 6 & 2\ 7 & 8 & 9 & 4\endarray ight> ight) = left<eginarrayccc 1 & 0 & 0 & frac-43\ 0 & 1 & 0 & frac83 \ 0 & 0 & 1 & -1\endarray ight>. $$

This implies the vector $vecx$ we seek is given by:$$vecx = left<eginarrayc frac-43\ frac83 \ -1\ endarray ight>.$$

The visibility of this $vecx$ alone assures $vecb in hedesigningfairy.comrmSpan(A)$, but lets examine our prize by computer $Avecx$.

See more: Which Item Among The Following Is Not An Intangible Asset? ? A

$$Avecx = left<eginarrayccc 1 & 2 & 3\ 4 & 5 & 6\ 7 & 8 & 9 endarray ight> left<eginarrayc frac-43\ frac83 \ -1\ endarray ight> = left<eginarrayc 1\ 2 \ 4\ endarray ight>, $$

as desired.

Some points to think about: are there any vectors in $ hedesigningfairy.combbR^3$ that room not in Span($A$)? If so can you uncover them, if not can you justify it? hope this price helps!