Finding a constant such that system is consistent

[TranscendArcu]

Homework Statement

The Attempt at a Solution

So I was thinking of trying to do row reduction in the hopes that would lead me to an answer.

$\left| \begin{array}{ccc}<br /> 0& 1& 1& 2 \\<br /> 1&1&1&a \\<br /> 1&1&0&2 \end{array} \right|$ → $\left| \begin{array}{ccc}<br /> 1&1&1&a \\<br /> 0&1&1&2 \\<br /> 1&1&0&2 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&1&1&2 \\<br /> 1&1&0&2 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&1&1&2 \\<br /> 0&1&0&-a+4 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&0&1&a-2 \\<br /> 0&1&0&-a+4 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&1&0&-a+4 \\<br /> 0&0&1&a-2<br /> \end{array} \right|$

So this seems to be telling me that I can choose any $a$ and $x=z=a-2$ and $y=-a+4$. Am I doing this right?

 

[SammyS]

Homework Statement

The Attempt at a Solution

So I was thinking of trying to do row reduction in the hopes that would lead me to an answer.

$\left| \begin{array}{ccc}<br /> 0& 1& 1& 2 \\<br /> 1&1&1&a \\<br /> 1&1&0&2 \end{array} \right|$ → $\left| \begin{array}{ccc}<br /> 1&1&1&a \\<br /> 0&1&1&2 \\<br /> 1&1&0&2 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&1&1&2 \\<br /> 1&1&0&2 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&1&1&2 \\<br /> 0&1&0&-a+4 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&0&1&a-2 \\<br /> 0&1&0&-a+4 <br /> \end{array} \right|$→$\left| \begin{array}{ccc}<br /> 1&0&0&a-2 \\<br /> 0&1&0&-a+4 \\<br /> 0&0&1&a-2<br /> \end{array} \right|$

So this seems to be telling me that I can choose any $a$ and $x=z=a-2$ and $y=-a+4$. Am I doing this right?

That all looks fine !