Find values of x so that the matrix is not invertible Given 
$$A=\begin{pmatrix}
5 &  6 &  6-x & 8 \\
2 &  2-x &  2 &  8 \\
6 &  6 &   2 &  8 \\
2 &  3 &   6 &  7
\end{pmatrix}\in\mathbb{R}^{4\times 4}$$
One must find all values of $x\in\mathbb{R}$ such that the matrix is not invertible.
I tried finding the determinant of $A$ and in the process got that $x = 4$ and $x = 6/192$. Therefore, $A1 = 1360; A2 = 816; A3 = 1146; A4 = 1088$. However when I calculate $\det(A) = 1360 - 816 + 1146 - 1088= 602$ and I do not get 0 so that I could prove it is not invertible. 
 A: HINT
Let's calculate $\det A$ as a function of x and then set $\det A=0$.
Note that adding a scalar multiple of one column/row to another column/row does not change the value of the determinant. Thus you can simplify the matrix A before to perform the calculation.
You should obtain:
$$\det A=2(-13 x^2 + 19 x - 4)$$
and from here find the values for which $\det A=0$.
Notably
$$\det A=\begin{vmatrix}
5 &  6 &  6-x & 8 \\
2 &  2-x &  2 &  8 \\
6 &  6 &   2 &  8 \\
2 &  3 &   6 &  7
\end{vmatrix}=
\begin{vmatrix}
-1 &  x &  6-x & 8 \\
x &  -x &  2 &  8 \\
0 &  4 &   2 &  8 \\
-1 &  -3 &   6 &  7
\end{vmatrix}=\\=-1\begin{vmatrix}
  -x &  2 &  8 \\
  4 &   2 &  8 \\
  -3 &   6 &  7
\end{vmatrix}-x\begin{vmatrix}
 x &  6-x & 8 \\
 4 &   2 &  8 \\
-3 &   6 &  7
\end{vmatrix}+1\begin{vmatrix}
  x &  6-x & 8 \\
 -x &  2 &  8 \\
 4 &   2 &  8 \\
\end{vmatrix}=\\=
-1\cdot(34x+136)-x\cdot(18x-72)+(128-8x^2)=-26x^2+38x-8 $$
A: Some properties of square matrices that may be useful in your case:  
If two rows/columns of a matrix are proportional the matrix is $0$. For example:
$$\begin{pmatrix}
a &  b &  c & d \\
e &  f &  g & h \\
x &  y &  z & w \\
ka & kb & kc & kd
\end{pmatrix}=0$$ 
You can multiply any row/column of a matrix with a constant and add it to another row/column. For example:
$$\begin{pmatrix}
a &  b &  c & d \\
e &  f &  g & h \\
x &  y &  z & w \\
i & j & m & n
\end{pmatrix}=\begin{pmatrix}
a &  b &  c & d+kb \\
e &  f &  g & h+kf \\
x &  y &  z & w+ky \\
i & j & m & n+kj
\end{pmatrix} $$ 
If any row/column of a matrix is all $0$, then the matrix is $0$. For example:
$$\begin{pmatrix}
a &  b &  c & d \\
e &  f &  g & h \\
0 &  0 &  0 & 0 \\
i & j & m & n
\end{pmatrix}=0$$ 
By using the second property multiple times on multiple rows and columns you can get one of the rows/columns to be $0$ or be proportional and then you get that $detA=0$.
