Q:

Number of real values of $x$ satisfying the following determinant?

The number of real values of $x$ which make the following determinant equal to $0$ are ?
$$
        \text{det}\left(\begin{matrix}
        x & 3x + 2 & 2x-1 \\
        2x-1 & 4x & 3x+1 \\
        7x-2 & 17x+6 & 12x-1 \\
        \end{matrix}\right)
$$
Answer: Infinite number of values
This is a homework question. I tried it but it became a bit lengthy. I tried it using various row and column operations on the determinant. At the end I found that $x$ was eliminated and thus I concluded that the value is independent of $x$. Is there a better way to do this ?

A:

Elementary row operations are indeed the best way to find the discriminant of an arbitrary matrix. I don't know why you say it is lengthy.
$\pmatrix{ x & 3x+2 & 2x-1 \\
2x-1 & 4x & 3x+1 \\
7x-2 & 17x+6 & 12x-1 }$
becomes:
$\pmatrix{ x & 3x+2 & 2x-1 \\
-1 & -2x-4 & -x+3 \\
-2 & -4x-8 & -2x+6 }$
which is obviously singular.

