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I've been trying to find the way to solve this problem.

Now based on what I've read so far that if $(k A)^{-1} = 1/k A^{-1}$

Since we already have the inverse of $(2A)^{-1}$ do I need to create an I?

Zgames55
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    HINT: $((2A)^{-1})^{-1} = 2A$. – Arturo Magidin Jan 08 '20 at 21:23
  • Yes thank you, I'm a bit of a slow learner, might take some time wrap around this. But the first comment and yours has helped me understanding. Also does this mean if I need to find $A$ I only need to divide the $2A$ with 2? – Zgames55 Jan 08 '20 at 21:42

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