<s>[INST] Find values for x such that A is not invertible. For $A$ not to be invertible, i.e. $\det(A) = 0$
$A =$ $$
    \begin{bmatrix}
    3 & 1 & 7-x \\
    3 & 2-x & 4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
    $$
After some row operations (step 1: R1 - R2 => R1 then R2 - R3 => R2 and lastly R1+R2 => R1) I got:
$A =$ $$
    \begin{bmatrix}
    -1 & -1 & 7-x \\
    -1 & -x & -4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
    $$
I computed the determinant.
Then checked when it equals $0$ and for values of $x$ I got    $\dfrac  {1}{4} (19\pm \sqrt{305})$.
Could you please confirm with me whether I am on the right track? Thanks.
 [/INST] I obtain $$detA=-4 x^2 + 6 x + 2=0\implies x=\frac34\pm \frac{\sqrt {17}}4$$
To find the values for which $det A=0$, you could simplify the matrix as follow
$$\begin{bmatrix}
    3 & 1 & 7-x \\
    3 & 2-x & 4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
\to\begin{bmatrix}
    3 & 1 & 1-x \\
    3 & 2-x & -2 \\
    2 & 1 & 0 \\
    \end{bmatrix}
$$
</s>