<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] "Subtracting a row of 2s" is NOT a valid matrix operation.  But subtracting one row from another is.  Perhaps what your teacher did (or meant to do) is, first subtract the third row from the second to get 
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 2 & 2 & x & 2 \\ 2 & 2 & 2 & x \end{array}\right|$.
Now, subtract the fourth row from the third to get 
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 0 & 0 & x- 2 & 2-x \\ 2 & 2 & 2 & x \end{array}\right|$.
Finally, you can subtract the first row from the fourth to get 
$\left|\begin{array}{ccc} x & 2 & 2 & 2 \\ 0 & x- 2 & 2- x & 0 \\ 0 & 0 & x- 2 & 2-x \\ 2- x & 0 & 0 & x- 2 \end{array}\right|$.
That's almost an "upper triangular" matrix.  We can calculate the determinant reasonably easily by "expanding on the first column:
$x\left|\begin{array}{cc} x- 2 & 2- x & 0 \\ 0 & x- 2 & 2- x \\ 0 & 0 & x- 2 \end{array}\right|$$- 2\left|\begin{array}{cc} 2 & 2 & 2 \\ x- 2 & 2- x & 0 \\ 0 & x- 2 & 2- x \end{array}\right|$.
That first determinant is $x(x- 2)^3$.  For the second we can further expand that three by three determinant on the first column to get $-2\left(2\left|\begin{array}{cc}2- x & 0 \\ x- 2 & 2- x\end{array}\right|- (x- 2)\left|\begin{array}{cc}2 & 2 \\ x- 2 & 2- x\end{array}\right|\right)= -4(2- x)^2- (x- 2)(4- 2x- 2x+ 4)= -4(2- x)^2- (x- 2)(8- 4x)= -4(2- x)^2+ 4(2- x)^2= 0$.
So the determinant of the given matrix is $x(x- 2)^3$.
</s>