# Matrix inversion with variable in {-1,1}

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 $\{1,-1\}$

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

Done using augmented matrix: $\left(\begin{array}{ccc}1&-fi&-g-h\sqrt2+fij\\0&i&-ij\\0&0&1\end{array}\right)$
• Could it be you thought $\;i=\sqrt{-1}\;$ ...just as I did the first time? :) – DonAntonio Jan 5 '14 at 17:12
• $i^2=1$ or $i=\frac1i$ anyway... – peterwhy Jan 5 '14 at 17:12
• That's true, @peter, yet $\;\frac1i=-i\;$ if $\;i=\sqrt{-1}\,,\,-1\;$ , but not if $\;i=1\;$ ... – DonAntonio Jan 5 '14 at 17:14
$$\begin{pmatrix} 1 & f & g+h\sqrt(2) \\ 0 & i & j \\ 0 & 0 & 1 \\ \end{pmatrix}^{-1}= \frac1i \begin{pmatrix} i & -f & \;\;fj-ig-ih\sqrt(2) \\ 0 & \;\;1 & \!-j \\ 0 & \;\;0 & \;\;i \\ \end{pmatrix}$$