For what values of $x$ does $A^{-1}$ exist? Let $A = \begin{bmatrix}4&-1&2\\5&x&7\\x&-1&3\end{bmatrix}$
I'm trying to find the values of $x$ such that $A^{-1}$ exists.
What I have tried:
$4(3x + 7) - (-1)(15)(7x) + (2)(-5)(x^2) = 0$
$-10x^2 + 117x + 28 = 0$
but I don't think that's right! 

Note that 
\begin{align*}
\det A&=4\cdot\left(3x+7\right)-(-1)\cdot\left(15-7x\right)+2\cdot\left(-5-x^2\right)\\
&=-2x^2+5x+33 \\
&=-2(x+3)\left(x-\frac{11}{2}\right)
\end{align*}
Hence $A$ is invertible if and only if $x\neq -3$ and $\displaystyle x\neq\frac{11}{2}$.
