# Find $x$ so that $\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible

Find $x$ so that the matrix
$\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible. Please show your work.

Let
$A=\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}.$
Then $\frac{1}{a}R_1+R_3 \rightarrow R_3$ gives
$\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ 0 & x & x^2+\frac{c}{a} \end{pmatrix},$
and $-\frac{x}{a}R_2+R_3 \rightarrow R_3$ gives
$\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ 0 & 0 & \frac{b}{a}x+x^2+\frac{c}{a} \end{pmatrix}.$
Hence
$\text{det}(A)=a(\frac{b}{a}x+x^2+\frac{c}{a})=0$
$\Rightarrow ax^2+b x+c=0$
$\Rightarrow x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$
Hence the matrix $A$ is invertible if and only if
$x\neq\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$