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

Question

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 ?

Answer 1

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.

Answer 2

If you add $3$ times the first row with $2$ times the second row, you end up with the third row.

Thus the rows are linearly dependent, hence the matrix is singular for all values of $x$, so that the determinant is $0$ for all values of $x$ as well.

Answer 3

Yes there is. It is the case that if your rows or columns are not linearly dependent then your determinant is zero. Given that is true about columns you can make the statement about rows since $\textrm{det}(A^T) = \textrm{det}(A)$. So observe the third rows is twice the second plus three times the first.