0
$\begingroup$

Let $ A=\begin{pmatrix} 2 & 1 &1 \\ 0 & 2 & 1\\ 0& 0& 2 \end{pmatrix}$ and $ B=A^{n}+A^{n-1}+...+A+I_{3} $.
Determine $ B^{-1} $.

So far I found that $ B=\sum_{k=0}^{n}\begin{pmatrix} 2^{k} & k\cdot 2^{k-1} &k\cdot (k+3)\cdot 2^{k-3} \\ 0 & 2^{k} &k\cdot 2^{k-1} \\ 0 & 0 & 2^{k} \end{pmatrix} $.

  • 2
    Have you seen sequences/series of the form $a^n+a^{n-1}+\cdots+a+1$ before? – 2017-01-04

1 Answers 1

2

$$B=(A^n-I)(A-I)^{-1}$$

Can you see why $\;A\;$ and $\;A-I\;$ are invertible? Also, can you see why $\;(A-2)^3=0\;$

  • 0
    I understand that $ B=(A^n-I)(A-I)^{-1} $ and $ detA=\prod_{i=1}^{3}a_{ii} $, but I don't know how to go on with the first relation. – 2017-01-04
  • 0
    @ztefelina What "first relation"? – 2017-01-04
  • 0
    $ B=(A^{n+1}-I)(A-I)^{-1} $ – 2017-01-04
  • 0
    @ztefelina That comes from the sum of a geometric series: $$a+ar+ar^2+\ldots+ar^n=a\frac{r^{n+1}-1}{r-1}$$ IN our case we're talking of matrices and thus, instead of dividing by $\;A-I\;$ we multiply by its inverse...that's all! – 2017-01-04
  • 0
    I know that what you've said is true, but how can I get its inverse from that one? – 2017-01-04
  • 0
    @ztefelina What do you want its inverse for?? – 2017-01-04
  • 0
    That's what I have to find :) – 2017-01-04
  • 0
    @ztefelina No, that is *not* what is written in your post....there it is written that you have to find $\;B^{-1}\;$ ... – 2017-01-04
  • 0
    For a square matrix A, the inverse is written $A^{-1}$. – 2017-01-04
  • 0
    @ztefelina I know that...so? – 2017-01-04
  • 0
    So, when I wrote "Determine $ B^{-1} $ I meant "Determine its inverse" :) – 2017-01-04
  • 0
    @ztefelina I know that, too...what is going on? You wrote above that you wanted to know $\;A^{-1}\;$ and I told you that you asked about $\;B^{-1}\;$ ... – 2017-01-04
  • 0
    No, I want to know $ B^{-1} $..$ A^{-1} $ is obvious. – 2017-01-04
  • 0
    @ztefelina I know...and after my answer also $\;B^{-1}\;$ is obvious. Look at the first line in my answer. – 2017-01-04
  • 0
    I do understand it now...sorry for chatting! :) – 2017-01-04