Find values of $x$ such that at least one of the eigenvalues of the matrix is zero

Question

Given the matrix

$$\begin{pmatrix} 0 & x & 5 & -10 \\ -x & 10 & -x & 10 \\ -10 & -x & 10 & -5 \\ 10 & -5 & -10 & -x \end{pmatrix}.$$

I need to choose values of $x \in [-16,0]$ such that at least one of the eigenvalues of the matrix is zero.

Answer 1

For a zero eigenvalue you need the determinant to be zero.

If you add the 3rd row to the 4th row you obtain $$0=\begin{vmatrix}0&x&5&-10 \\-x&10&-x&10\\-10&-x&10&-5\\0&-5-x&0&-5-x\\\end{vmatrix}$$ Therefore either $x=-5$ or $$0=\begin{vmatrix}0&x&5&-10 \\-x&10&-x&10\\-10&-x&10&-5\\0&1&0&1\\\end{vmatrix}.$$ Now add multiples of Row 4 to the other rows to obtain $$0=\begin{vmatrix}0&10+x&5&0 \\-x&0&-x&0\\-10&5-x&10&0\\0&1&0&1\\\end{vmatrix}=\begin{vmatrix}0&10+x&5 \\-x&0&-x\\-10&5-x&10\\\end{vmatrix}=-x\begin{vmatrix}0&10+x&5 \\1&0&1\\-10&5-x&10\\\end{vmatrix}$$

Therefore $x$ is $-7,-5$ or $0$.