Finding missing row of $3\times 3$ matrix given desired eigenvalues of the matrix Given matrix
$$
C=
\begin{bmatrix}
  0 & 1 & 0 \\
  0 & 0 & 1 \\
  x & y & z
\end{bmatrix}
$$
I want to put in the final row that makes it so the eigenvalues of the matrix are $1,2,3$.
I know that the trace of C should equal the sum of the eigenvalues. Therefore
$$z=6$$
I know that the determinant must be equal the the eigenvalues multiplied together. $\text{Det}(C) = x$. So now I know
$$x = 6xy \implies y = 1/6$$
I'm left with $x$ being anything but clearly it can't be anything... What's the appropriate way to solve for the last variable?
 A: $C$ is the transpose of a companion matrix. Therefore, the last row consists of the negative coefficients of the characteristic polynomial. The characteristic polynomial is
$$
\chi(t) = (t-\lambda_1)(t-\lambda_2)(t-\lambda_3)
= (t-1)(t-2)(t-3) = t^3-6t^2+11t-6
$$
from which we can derive $x=6,$ $y=-11$ and $z=6.$
A: Two of your observations are correct, but you mix up elements and eigenvalues in some places which gives some faulty equations.
$z=e_1+e_2+e_3=1+2+3=6$ equals trace is a reasonable observation
$x=det = e_1\cdot e_2\cdot e_3=1\cdot 2 \cdot 3=6$ also.
We have only $y$ left to determine. I think some relatively straight forward brute force calculation should be able to find $y$ if it exists.
Let us look at the easiest eigenvalue $= 1$. It means there exists 1 non-zero vector that will be taken to the 0 vector by the matrix:
$$\begin{bmatrix}-1&1&0\\0&-1&1\\6&y&5\end{bmatrix}$$
Let's assume it is $[a,b,c]^T$
now $[b-a,c-b,6a+yb+5c]=[0,0,0]$
Already the first two equations say $b=a, c=b$ which tells us that they must all be the same.
$6+y+5=0$ gives $y = -11$.
We can verify that it works. For example in Gnu Octave with the code:
eig([0,1,0;0,0,1;6,-11,6])'

gives us $[1,2,3]$.
But it is of course a good exercise to do it by hand.
A: You need $C - \lambda I$ to be singular for $\lambda = 1,2,3$. Just compute these matrices with $z=6$ and $y = 1/6$ and one of them should give you a constraint on $x$.
A: Compute the characteristic polynomial of the matrix. It has three terms and you may notice that two of the coefficients are the trace and the determinant (up to some sign). The third coefficient equals 12 + 23 + 3*1 or in general the sum of the three products of three eigenvalues.
