<s>[INST] Determine all real $x$ that satisfy $\det A=0$ I want to find all real $x$ that satisfy 
$$
  \textrm{det } X=
  \begin{vmatrix}
  x &2 &2 &2\\
  2 &x &2 &2\\
  2 &2 &x &2\\
  2 &2 &2 &x
  \end{vmatrix}\\
$$
My teacher does this by adding the three bottom rows to the top row
$$
\textrm{det } X=
  \begin{vmatrix}
  x+6 &x+6 &x+6 &x+6\\
  2 &x &2 &2\\
  2 &2 &x &2\\
  2 &2 &2 &x
  \end{vmatrix}\\
$$
and then subtracting a row of $2$'s from the bottom three rows
$$
\textrm{det } X=
  (x+6)  
  \begin{vmatrix}
  1 &1 &1 &1\\
  0 &x-2 &0 &0\\
  0 &0 &x-2 &0\\
  0 &0 &0 &x-2
  \end{vmatrix}.
$$
The answer is 
$$
x\in \{-6,2\}.
$$
I think I understand the operations (although subtracting an arbitrary row of numbers from a matrix/determinant row is something I've never seen before, but I don't see why that wouldn't be allowed. Just like you can subtract arbitrary coefficients on both sides of an equation, right?), my main issue is why they are performed. 


*

*Why can't I just in the same way subtract a row of $2$'s from the three bottom rows in the first determinant? If I do that I get a different answer. 

*I know I want a column of all zeroes except one column-element, but why do I need to perform the first operation beforehand? Is it somehow necessary that all the top row elements to be the same, $(x+6)$?

 [/INST] Another way: the determinant is a fourth-degree polynomial $p(x)$ in variable $x$.
It is easy to see that $p(2) = p(-6) = 0$, like you did.
Then take the derivative:
\begin{align}
p'(x) &= \begin{vmatrix}
  1 &2 &2 &2\\
  0 &x &2 &2\\
  0 &2 &x &2\\
  0 &2 &2 &x
  \end{vmatrix} + \begin{vmatrix}
  x &0 &2 &2\\
  2 &1 &2 &2\\
  2 &0 &x &2\\
  2 &0 &2 &x
  \end{vmatrix} + \begin{vmatrix}
  x &2 &0 &2\\
  2 &x &0 &2\\
  2 &2 &1 &2\\
  2 &2 &0 &x
  \end{vmatrix} + \begin{vmatrix}
  x &2 &2 &0\\
  2 &x &2 &0\\
  2 &2 &x &0\\
  2 &2 &2 &1
  \end{vmatrix} \\
&=4  \begin{vmatrix}
  x &2 &2\\
  2 &x &2\\
  2 &2 &x\\
  \end{vmatrix} \\
\end{align}
Taking derivatives again we get $p''(x) = 4\cdot 3 \begin{vmatrix}
  x &2 \\
  2 &x \\
  \end{vmatrix}$ so we can conclude that $p'(2) = p''(2) = 0$.
Therefore $2$ is a root of $p$ with muliplicity $3$ so $$p(x) = (x-2)^3(x+6)$$
Therefore the zeros are indeed only $-6$ and $2$.
</s>