1
$\begingroup$

Could you please give me a hint for computing inversion of this matrix?

$$ \begin{pmatrix} 1 & f & g+h\sqrt(2) \\ 0 & i & j \\ 0 & 0 & 1 \\ \end{pmatrix} $$

where $f,j \in \mathbb Z;g,h\in \mathbb Q;i\in \{-1,1\}$

I can't use this formula:

$$ \displaystyle (A^{-1})_{ij}=(-1)^{i+j}\frac{\mathop{\rm det}\nolimits A_{j,i}}{\mathop{\rm det}\nolimits A}\, $$

I'm getting this matrix: $$ \begin{pmatrix} 1 & f & 0 \\ 0 & i & j \\ 0 & 0 & 1 \\ \end{pmatrix} $$

but don't know what to do next, as $i$ is $\{1,-1\}$

$\endgroup$
  • 1
    $\begingroup$ Do you mean "the inverse of the matrix", or inversion of a matrix in another meaning? $\endgroup$ – DonAntonio Jan 5 '14 at 17:05
  • 1
    $\begingroup$ Yes, the inverse, sorry. $\endgroup$ – DropDropped Jan 5 '14 at 17:07
  • $\begingroup$ And why can't you use that formula which, in fact, is the famous formula with the adjoint of $\;A\;$ ? $\endgroup$ – DonAntonio Jan 5 '14 at 17:08
  • $\begingroup$ Have you tried using augmented identity matrix? $\endgroup$ – peterwhy Jan 5 '14 at 17:08
1
$\begingroup$

Done using augmented matrix: $\left(\begin{array}{ccc}1&-fi&-g-h\sqrt2+fij\\0&i&-ij\\0&0&1\end{array}\right)$

$\endgroup$
  • $\begingroup$ Could it be you thought $\;i=\sqrt{-1}\;$ ...just as I did the first time? :) $\endgroup$ – DonAntonio Jan 5 '14 at 17:12
  • $\begingroup$ $i^2=1$ or $i=\frac1i$ anyway... $\endgroup$ – peterwhy Jan 5 '14 at 17:12
  • 1
    $\begingroup$ That's true, @peter, yet $\;\frac1i=-i\;$ if $\;i=\sqrt{-1}\,,\,-1\;$ , but not if $\;i=1\;$ ... $\endgroup$ – DonAntonio Jan 5 '14 at 17:14
0
$\begingroup$

$$ \begin{pmatrix} 1 & f & g+h\sqrt(2) \\ 0 & i & j \\ 0 & 0 & 1 \\ \end{pmatrix}^{-1}= \frac1i \begin{pmatrix} i & -f & \;\;fj-ig-ih\sqrt(2) \\ 0 & \;\;1 & \!-j \\ 0 & \;\;0 & \;\;i \\ \end{pmatrix} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.