Solution set for $Ax=\bf b$, with a parameterized matrix over the complex numbers

$\mathbf{F}=\mathbb{C}:$ $$\left(\begin{array}{ccc|c} 1 &\lambda -2&0&0\\ \lambda +2&-5&0&0\\ 0&0&1&1 \end{array}\right)$$

For which values of $\lambda$ the system has:

1. Unique solution
2. No solution

Edit: I've reached that, but don't know if it has any significance: $$\left\{ \begin{array}{l}x_1 +(\lambda -2)x_2 = 0\\ (\lambda+2)x_1-5x_2 = 0 \end{array} \right.$$ $$-(\lambda-2)(\lambda+2)-5=0$$ $$-\lambda^2+4-5=0$$ $$\lambda=\sqrt{-1}=i$$

Edit: I now know that if $\lambda=i$ there are families of answers, and unique answers if $\lambda\not=i$. Is there a way to reach this using only elementary row operations?

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I guess the first thing one should note is that this consists of two systems, one is $x_3=1$ and the other is a 2-by-2 homogeneous system. – Hui Yu Nov 10 '12 at 16:07
That matrix in the title really annoyed me. $3\times 3$ is too big for the title, and you should probably avoid it in the future ($2\times 2$ is borderline reasonable). – Asaf Karagila Nov 10 '12 at 16:52
Thank you for the comment, asaf. I will. – yuvalz Nov 10 '12 at 16:56

Note that the system will always have at-least one solution since $(0,0,1)$ satisfies the linear system irrespective of the value of $\lambda$. Hence, the question is whether the system has a unique solution or infinite solution. Look at the determinant of the matrix $$A = \begin{pmatrix} 1 &\lambda -2&0\\ \lambda +2&-5&0\\ 0&0&1 \end{pmatrix}$$

Move your mouse over the gray area to get the value of the determinant.

$\det(A) = -5 - (\lambda-2)(\lambda+2) = - 5 -(\lambda^2-4) = - (1+\lambda^2)$.

If $\det(A) = 0$, then the system will have infinite solutions.

If $\det(A) \neq 0$, then the system will have a unique solutions.

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Is there a way to get this without using determinants? We haven't learnt them yet. Should use "elementary functions" (I think that's what they are called). Like adding/subtracting/multiplying lines. – yuvalz Nov 10 '12 at 16:25
I now see how my the "i" I've reached in my edit satisfies your infinite solutions case. Question is, is there a way I can show that using only elementary row operations? – yuvalz Nov 10 '12 at 16:41
Found the way, thank you. – yuvalz Nov 10 '12 at 16:49