Find the condition on $a,b,c$ such that $(a,b,c) \in \operatorname{span}((2,1,0) , (1,-1,2), (0,3,-4))$

Question

Find the condition on $a$, $b$, $c$ so that $(a,b,c) \in \mathbb{R^3}$ is in the space generated by $(2,1,0)$, $(1,-1,2)$ and $(0,3,-4)$

If $(a,b,c)$ lies in the space generated by given vectors then it must be equal to the linear span.

$(a,b,c) = \alpha(2,1,0) +\beta(1,-1,2) + \gamma(0,3,-4)$

Upon solving this system for $\alpha$, $\beta$ and $\gamma$ I ended up with following equations

$\alpha - \beta + 2\gamma = b$

$0\alpha + 3\beta - 4\gamma = a-2b$

$0\alpha + 0\beta + 0\gamma = c-a+2b$

Since the given system needs to be consistent we must have $c-a+2b = 0$

However, my book says the correct answer is $2a - 4b - 3c = 0$

Please tell me what is wrong with my solution and what should be the correct way to solve it.

Thank you.

Answer 1

Your book is right. The system is$$\left\{\begin{array}{l}2\alpha+\beta=a\\ \alpha-\beta+3\gamma=b\\ 2\beta-4\gamma=c.\end{array}\right.$$Multiplying the second equation by $-2$ and adding this to the first one, you get$$\left\{\begin{array}{l}3\beta-6\gamma=a-2b\\ \alpha-\beta+3\gamma=b\\ 2\beta-4\gamma=c.\end{array}\right.$$This is equivalent to$$\left\{\begin{array}{l}\beta-2\gamma=\frac13(a-2b)\\ \alpha-\beta+3\gamma=b\\ \beta-2\gamma=\frac12c\end{array}\right.$$So, the system is consistent if and only if $\frac13(a-2b)=\frac12c$, which is equivalent to the answer provided by your book.

Answer 2

$(a,\color{blue}{b},c)$ = $\color{blue}{\alpha}(2,\color{blue}{1},0)$ +$\color{blue}{\beta}(1,\color{blue}{-1},2)$ + $\color{blue}{\gamma}(0,\color{blue}{3},-4)$

Upon solving this system for $\alpha$ , $\beta$ and $\gamma$ I ended up with following equations

$\alpha$ - $\beta$ + $\color{red}{2}\gamma$ = $b$

How did you arrive at the equation above? Looking at the second coordinate (in blue); I'd expect: $$b=\alpha-\beta+\color{purple}{3}\gamma$$

Answer 3

Let us try to find $c_1$, $c_2$, $c_3$ such that the given vectors fulfill the condition $c_1x_1+c_2x_2+c_3x_3=0$. (If this is true, any vector $(a,b,c)$ belonging to their span fulfills the same condition.)

From this we get a system of three linear equations \begin{align*} 2c_1+c_2&=0\\ c_1-c_2+2c_3&=0\\ c_1+2c_2-2c_3&=0 \end{align*}

Solving this system of equations we get that $(c_1,c_2,c_3)$ has to be a multiple of the vector $(2,4,3)$.

This can be said equivalently like this: The span of the three given vectors is exactly the set of vectors $\{(x_1,x_2,x_3)\in\mathbb R^3; 2x_1+4x_2+3x_3=0\}$.

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This is related to the fact that every subspace of $\mathbb R^n$ is equal to solution set of some homogeneous system of linear equations.

By solving the system with unknowns $c_{1,2,3}$ we found such homogeneous system. (In this case it consisted of only one equation.)