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I am finding inverse matrix $A^{-1}$ and I was given hint that I could firstly find inverse matrix to matrix B which is transformed from A. $$A=\begin{pmatrix}1 &3 & 9& 27\\3 & 3 & 9& 27\\ 9 & 9 & 9& 27 \\ 27 & 27 &27& 27\end{pmatrix}$$

$$B=\begin{pmatrix}1 &3 & 9& 27\\1 & 1 & 3& 9\\ 1 & 1 & 1& 3 \\ 1 & 1 &1& 1\end{pmatrix}$$

I found that that $B^{-1}$ is $$B^{-1}=\begin{pmatrix}-\frac{1}{2} &\frac{3}{2} & 0& 0\\\frac{1}{2} & -2 & \frac{3}{2}& 0\\ 0 & \frac{1}{2} & -2& \frac{3}{2} \\ 0 & 0 &\frac{1}{2}& -\frac{1}{2}\end{pmatrix}$$

Now I don't know how to continue. What rule do I use to find $A^{-1}$?

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Note that $A=\begin{pmatrix}1 &0 & 0& 0\\0 & 3 &0 &0 \\ 0 & 0 & 9& 0 \\ 0 & 0 &0& 27\end{pmatrix}B$. Hence, $A^{-1}=B^{-1}\begin{pmatrix}1 &0 & 0& 0\\0 & 1/3 &0 &0 \\ 0 & 0 & 1/9& 0 \\ 0 & 0 &0& 1/27\end{pmatrix}$

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  • $\begingroup$ Thanks, I solved it correctly according to solution in my textbook :). $\endgroup$ – cgnx Jun 7 '14 at 23:11

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