For what values of $k$ does the determinant of this matrix 'vanish?'

Question

\begin{bmatrix} k & 1 & 4 \\ 1 & k & 3 \\ 1 & 0 & 1 \end{bmatrix}

So I think to co-factor expand along the 3rd row giving me

$$1(3-4k)-0+1(k^2-1)$$

Which I guess can be solved to a weird quadratic that doesn't seem to get me anywhere. I was doing this because I thought this would reveal what k give me a determinant equal to $0$. Which would answer my question.

Do I need to perform row operations beforehand? Or is there another method to finding the values of k to satisfy the question?

Thanks in advance, tips/hints/solutions are appreciated :)

Answer 1

Your approach is fine.

$$k^2+3-4k-1=0$$

$$k^2-4k+2=0$$

Use the formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Answer 2

You're going in entirely the right direction. It's not a weird quadratic at all. It's just that the unknown is called $k$ instead of $x$, and you need to rearrange it a little and set it equal to $0$ to get it on the standard form.