What is the smallest value of $x$ such that the matrix $A$ is not invertible?

Question

$$A = \begin{pmatrix} -1 & -3 & x \\ -2 & -7 & -1 \\ x & -6 & 1 \end{pmatrix}$$

So far I tried taking the determinant by expanding down the last column but i still cant get the right answer.

x * det $ \begin{pmatrix} -2 & -7 \\x & -6 \end{pmatrix} $ + 1 * det $ \begin{pmatrix} -1 & -3 \\ x & -6 \end{pmatrix} $ +1*det $ \begin{pmatrix} -1 & -3 \\ -2 & -7 \end{pmatrix} $

I end up with $7x^2+15x+7$$=$$0$

Answer 1

You're on the right way continue by finding the roots

$x_1=-\frac{15}{14} - \frac{\sqrt{29}}{14}$

$x_2=-\frac{15}{14} + \frac{\sqrt{29}}{14}$

And taking the $\min\{x_1,x_2\} = -\frac{15}{14} - \frac{\sqrt{29}}{14}$

That's it that is your answer