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