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**1. Homework Statement**

Find out for which values of ##\alpha## the system ##AX + \alpha X = (1, 1, 1, 1)^t## has solutions.

$$A = \begin{pmatrix}

6 & 0 & -1 & 2 \\

3 & 5 & -3 & 6 \\

-2 & 0 & 7 & -4 \\

2 & 0 & 1 & 0

\end{pmatrix}

X = \begin{pmatrix}

x \\

y \\

z \\

t

\end{pmatrix}$$

**2. Homework Equations**

**3. The Attempt at a Solution**

At first I wanted to proceed with a system, since it's what the problem asks. So I wrote ##AX + \alpha X = (1, 1, 1, 1)^1## like this:

##\begin{cases}

6x - z + 2t + \alpha x = 1 \\

3x + 5y - 3z + 6t + \alpha y= 1 \\

-2x + 7z - 4t + \alpha z= 1\\

2x + z + \alpha t = 1

\end{cases}##

And then I should proceed, right?

But the professor showed me that you can do this with matrices this way too:

##AX + \alpha X = (A + \alpha I_4)X##

And that if the determinant of ##(A + \alpha I_4) \neq 0 \implies \exists! \alpha##, while if it's equal to 0 the solution depends from ##rank(A + \alpha I_4)## and ##rank(A + \alpha I_4 | B)##, which I don't understand what the professor meant with ##B##. Would it be the ##(1, 1, 1, 1)^t##? In this case, if the two ranks are equal, does it mean it still has more than just one solution? And if they are not equal, does it mean no solution at all?

Anyway, I start by finding the determinant of ##(A + \alpha I_4)## but I find myself with a ##4^{th}## grade equation like this: ##\alpha^4 + 18 \alpha^3 + 105 \alpha^2 + 208 \alpha + 40##

I'd like to know if someone finds out a different determinant, because it could be that I messed up while calculating it.