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Let $W\subset \mathbb R^3$ is the set of all $(a,b,c)$ such that the system $$x+2y+3z=a$$ $$4x+5y+6z=b$$ $$7x+8y+9z=c$$

is consistent. Prove that $W$ is the vector subspace of the vector space $\mathbb R^3$.

System is consistent if $rank(A)=rank(B)$ where $$ A= \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{bmatrix}\Rightarrow rank(A)=2$$ and $$B=\begin{bmatrix} 1 & 2 & 3 & a\\ 4 & 5 & 6 & b\\ 7 & 8 & 9 & c\\ \end{bmatrix}$$ $rank(B)=2$ if $a-2b+c=0$.

Now we need to show that $W$ is the vector subspace of the vector space $\mathbb R^3$.

$1.$ Closure of $W$ under addition

$$\forall x,y\in W,x+y\in W$$

$2.$ Closure of $W$ under scalar multiplication in $\mathbb R^3$ $$\forall x\in W,\forall\alpha\in\mathbb R^3, \alpha x\in W$$

How to prove these properties?

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    $\begingroup$ You have shown that $(a,b,c) \in W$ provided that $a-2b+c=0$. So assume $x=(a_1,b_1,c_1)$ and $y=(a_2,b_2,c_2)$ are both in $W$. Can you show $x+y \in W$, i.e., $(a_1+a_2)-2(b_1+b_2)+(c_1+c_2)=0$? Similarly for showing $W$ is closed under scalar multiplication. $\endgroup$ – kccu Apr 19 '17 at 16:04
  • $\begingroup$ @kccu, What is the actual procedure for showing that $(a_1+a_2)-2(b_1+b_2)+(c_1+c_2)=0$ and closure under scalar multiplication? $\endgroup$ – user300045 Apr 19 '17 at 16:19
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    $\begingroup$ You know that $a_1-2b_1+c_1=0$ and $a_2-2b_2+c_2=0$ because $x,y \in W$. So you can actually compute the value of $(a_1+a_2)-2(b_1+b_2)+(c_1+c_2)$ by rearranging it. $\endgroup$ – kccu Apr 19 '17 at 16:35
  • $\begingroup$ @kccu, And what about the closure under scalar multiplication? We have $\alpha=(u,v,w),x=(a_1,b_1,c_1)\Rightarrow \alpha x=u(a_1+b_1+c_1)+v(a_1+b_1+c_1)+w(a_1+b_1+c_1)$. Is this correct? Now, how to show that $\alpha x\in W$? $\endgroup$ – user300045 Apr 19 '17 at 16:45
  • $\begingroup$ Condition 2 is wrong: it should read $$\forall x\in W,\forall\alpha\in\mathbb R, \alpha x\in W$$ $\endgroup$ – N74 Apr 19 '17 at 21:20
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The system is consistent if and only if the vector $(a,b,c)$ is in the image of the map defined by your matrix $A$. Therefore the set $W$ is precisely the image of that map. But the map defined by a matrix is a linear map, and the image of a linear map is always a vector subspace. Therefore the rank of $A$ is immaterial.

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You can show that any linear combinations $\vec{a}, \vec{b} \in W$ also admits to a solution that can be found easily and without using the rank-nullity theorem. If $\vec{a}$ and $\vec{b}$ are consistent for the given system $A$, then $$\underbrace{Ax}_{\vec{a}} + c\underbrace{Ay}_{\vec{b}} = \underbrace{A(x + cy)}_{\text{Consistent}} = \vec{a} + c\vec{b}$$ Where $c$ is some arbitrary scalar and $x + cy$ is the solution, so the subspace is closed under scalar multiplication and addition. Note that we assumed $\vec{a}, \vec{b} \in W$, so it must be that $\vec{x}, \vec{y}$ exists such that $Ax = \vec{a}$ and $Ay = \vec{b}$

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