How can I find $\displaystyle $t$$ for system of linear equations, that the system would not be solvable?

I tryed with this code:

Code:`x = 10`

y = 10

z = 10

Solve[{ 9 x - 8 y + t*z == 1, 6 x - 8 y - 5 z == 1, 3 x -7 y + 9 z == 1}, {t}]

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- Nov 5th 2010, 09:44 AMNforceNot solvable system of linear equations in Mathematica
How can I find $\displaystyle $t$$ for system of linear equations, that the system would not be solvable?

I tryed with this code:

Code:`x = 10`

y = 10

z = 10

Solve[{ 9 x - 8 y + t*z == 1, 6 x - 8 y - 5 z == 1, 3 x -7 y + 9 z == 1}, {t}]

- Nov 5th 2010, 09:50 AMAckbeet
You have a highly over-determined system there. There are no solutions. Think about it:

90-80+10t=1

60-80-50=1

30-70+90=1

Some of those equations are downright incorrect. I would review your steps for arriving at that system of equations. What's the bigger context for this problem? - Nov 5th 2010, 09:59 AMNforce
Sorry, forget x,y,z values.

I need to find t to be not solvable. For t = 1, it's solvable, for t = 2 it's solvable and so on...

but when it's not? - Nov 5th 2010, 10:01 AMAckbeet
Ok. Try setting

$\displaystyle \left|\begin{matrix}9&-8&t\\6&-8&-5\\3&-7&9\end{matrix}\right|=0.$

That's a determinant there. What do you get? - Nov 5th 2010, 10:03 AMAckbeet
Whoops. That approach won't work. Take a look at this thread. Make sure you scroll all the way down to the end to get the real reason why that works.

- Nov 5th 2010, 10:13 AMNforce
I get a solution. Great.

It works, your approach works. - Nov 5th 2010, 10:14 AMAckbeet
Wonderful. What is your value for t?

- Nov 5th 2010, 10:20 AMNforce
-(137/6)

- Nov 5th 2010, 10:22 AMAckbeet
That is correct.