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My doubt is related to Cramer's rule with this exercise. I need to get the parametric equations of $$x-y-z-1=0,$$ $$y-t-z=0.$$

I assign parameters as $z := \lambda, t := \beta$, so I get

$$x-y=1+\lambda,$$ $$y=2+\beta$$

Now, using Cramer's rule is

$$x = \frac{\begin{vmatrix} 1+\lambda & -1 \\ 2+\beta & 1 \end{vmatrix}}{\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix}} = \frac{1+\lambda-2-\beta}{1}= \lambda-1-\beta$$

But if we solve in a traditional way, $$x=y+1+\lambda=3+\lambda+\beta.$$

In fact, the first solution for $x$ doesn't satisfy the system of equations. However, I think that the Cramer's rule is right.

So, what is wrong here?

Thanks!

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It's because$$\begin{vmatrix}1+\lambda&-1\\2+\beta&1\end{vmatrix}=1+\lambda-(-1)\times(2+\beta)=3+\lambda+\beta.$$

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  • $\begingroup$ All right... I was blind! Thanks! $\endgroup$ – user183002 Nov 3 '19 at 10:57

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