-2

How to transfer the following matrix into Smith normal form? $$\left[\begin{matrix} 2 & -2b & 0 \\ 0 & 2 & -2c \\ -2a & 0 & 2 \end{matrix}\right]$$ The final answer is $$\left[\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2|abc-1| \end{matrix}\right]$$

But how we get it?

Bernard
  • 175,478
Rosa
  • 301
  • Remember that you might need to use both column and row operations when computing smith normal form, whereas one or the other suffices for Gaussian elimination over a field. – Badam Baplan May 28 '18 at 21:19

1 Answers1

2

\begin{align} &\begin{bmatrix} 2 & -2b & 0 \\ 0 & 2 & -2c \\ -2a & 0 & 2 \end{bmatrix}\xrightarrow{r_3\leftarrow r_3+ar_1} \begin{bmatrix} 2 & -2b & 0 \\ 0 & 2 & -2c \\ 0 & -2ab & 2 \end{bmatrix}\xrightarrow{c_2\leftarrow c_2+bc_1\;\:} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & -2c \\ 0 & -2ab & 2 \end{bmatrix} \\[1.5ex] {}\xrightarrow{r_3\leftarrow r_3+abr_2} &\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & -2c \\ 0 & 0 & 2-2abc \end{bmatrix} \xrightarrow{c_3\leftarrow c_3+cc_2} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2(1-abc) \end{bmatrix} \end{align} Possibly change the sign of the last row.

Bernard
  • 175,478
  • last operation isn't a row or column operation instead should it be something like $\frac{c}{1-abc}r_3+r_2 \to r_2$ as long as $abc \neq 1$ and $cc_2+c_3 \to c_3$ if $abc=1$ – N8tron May 28 '18 at 21:38
  • But In smith form division is not allowed @N8tron – Rosa May 28 '18 at 21:40
  • okay well I clearly didn't get that. So it should just be $cc_2+c_3 \to c_3$? Just a simply typo? – N8tron May 28 '18 at 21:41
  • 1
    Sorry, I copy-pasted from one step to another, and I didn't fully change the description of the elementary operation. 'Tis fixed now. – Bernard May 28 '18 at 21:46