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$ \begin{bmatrix} k_1 & 0 & 0 & 0 \\ 0 & k_2 & 0 & 0 \\ 0 & 0 & k_3 & 0 \\ 0 & 0 & 0 & k_4 \\ \end{bmatrix} $

I'm thinking you just have to convert this into a reduced-row echelon form like this:

$ \left[ \begin{array}{cccc|cccc} k_1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & k_2 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & k_3 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & k_4 & 0 & 0 & 0 & 1\\ \end{array} \right] $

So the result for the inverse would be this:
$ \begin{bmatrix} \frac1{k_1} & 0 & 0 & 0 \\ 0 & \frac1{k_2} & 0 & 0 \\ 0 & 0 & \frac1{k_3} & 0 \\ 0 & 0 & 0 & \frac1{k_4} \\ \end{bmatrix} $

Is this right, or is there something I'm missing?

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    $\begingroup$ It is right. To check, just multiply the matrices, you get the identity matrix. Well done. $\endgroup$ – user405743 Jun 25 '17 at 4:26
  • $\begingroup$ Ok, thanks @Dragon $\endgroup$ – Bucephalus Jun 25 '17 at 4:26
  • $\begingroup$ Thanks for the edit @Daichi $\endgroup$ – Bucephalus Jun 25 '17 at 4:34
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    $\begingroup$ Further exercise: generalize this result to an $n \times n$ matrix. $\endgroup$ – Duncan Ramage Jun 25 '17 at 4:52
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    $\begingroup$ As Duncan was saying, this is a general result that the inverse of a diagonal matrix (it it is invertible) is another diagonal matrix having entries reciprocal of the entries of the given matrix. $\endgroup$ – StubbornAtom Jun 25 '17 at 7:29
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Because the matrix is invertible we can assume that all the $k_i$'s are non-zero. Your answer is correct. Elegant and sweet.

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  • $\begingroup$ This should be a comment, not an answer. $\endgroup$ – user261263 Jun 27 '17 at 4:20

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