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I have the linear system: $$ \begin{bmatrix} 2 & -1 & 3 & 5\\ 4 & 2 & 2 & 6\\ -2 & \alpha & 3 & 1 \end{bmatrix} $$

I want to find all $\alpha$ such that the system has infinitely many solutions. From what I've seen, there must be two equations that are multiples of each other so that they cancel out. I don't see an $\alpha$ that could work, so is there no $\alpha$ such that the system has infinitely many solutions?

Also, for a $3 \times 3$ matrix, how do you generally find $\alpha$ such that it has infinitely many solutions? I've only found examples for $2 \times 2$ matrices.

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$$ \begin{pmatrix} 2 & -1 & 3 & 5\\ 4 & 2 & 2 & 6\\ -2 & \alpha & 3 & 1 \end{pmatrix}\longrightarrow \begin{pmatrix} 2 & -1 & 3 & 5\\ 0 & 4 & \!\!\!-4 & \!\!\!-4\\ 0 & \alpha-1 & 6 & 6 \end{pmatrix} $$

So how can you do to have one row linearly dependent on the other two?

Spoiler!

$\,\alpha=-5\,$

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