I need to find all invariant factor of matrix
$\begin{pmatrix} \lambda +1 & 2 & -1 \\ 1 & \lambda & -3 \\ 1 & 1& \lambda-4 \end{pmatrix} = A(\lambda)$
that is, by existence of the Smith Normal Form there exists 3×3 matrices $B, C$ with entries in the
integers such that
$BA(\lambda) C = \text{diag}\{d_1, d_2, d_3\}.$
I must determine $d_1, d_2, d_3$. Applying row-column operations I arrive at the matrix
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & -\lambda^2 -\lambda+2 & 3\lambda +2 \\ 0 & 1-\lambda& \lambda-1 \end{pmatrix}$
at this point I can't find a pivot because the absolute value of the entries which are not $0$ or $1$ in the latter matrix vary according to $\lambda$. This problem is found in Jacobson's Basic Algebra I, chapter 3, section 7, problem 3. The problem suggests the usage of the following theorem:
Let A be an m × n matrix with entries in a
p.i.d. D and suppose the rank of A to be r. For each i ≤ r let Δi
be a g.c.d. of the i-rowed minors of A, then any set of invariant factors for A differ by unit
multipliers from the elements $d_1 = \Delta_1, d_2 = \Delta_2 \Delta_1^{-1}, \dots, d_r = \Delta_r\Delta_{r-1}^{-1}$.
But this theorem wasn't covered in my lectures so I'm not supposed to use it, so I would like help on how to obtain them manually.
Best Answer
$$\begin {pmatrix}1& 1&\lambda-4\\0& \lambda-1&1-\lambda \\0& 1-\lambda &-1-(1+\lambda)(\lambda -4)\end {pmatrix}\to\begin {pmatrix}1& 0& 0\\0& \lambda-1& 1-\lambda \\0& 0& -\lambda-(1+\lambda)(\lambda -4)\end {pmatrix}\to \begin{pmatrix}1&0&0\\0&\lambda-1&0\\0&0&-\lambda-(1+\lambda)(\lambda -4)\end {pmatrix}$$
Finally there's a divisibility issue. To rectify it, see here.