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

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

  1. Unique solution
  2. No solution
  3. Infinite amount of answers

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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    Thank you for the comment, asaf. I will.2012-11-10

1 Answers 1

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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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    Found the way, thank you.2012-11-10