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Compute the product of the quaternions $(a+ bi +aj)(ai + ck)$ with $a,b,c\in \mathbb Z.$

But, answer shown on Wolfram Alpha is different?

jiten
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    What is the answer on Wolfram Alpha? Did you notice any assumptions like "assuming i is an imaginary unit"? In any case, the edited final line seems to be correct now. – Antoine Jul 12 '22 at 07:25
  • @Antoine Thanks for help. Request one more related issue: If not wrong, all $i,j,k$ are non-real quantities. Hence, for taking inverse of $1-2i+j$ get by conjugation $\frac{1+2i-j}{6}$? But, this answer is stated wrong. – jiten Jul 12 '22 at 07:29
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    The inverse of $1-2i+j$ is indeed $\dfrac{1+2i-j}{6}$. See this result. – Zerox Jul 12 '22 at 07:32
  • @Zerox Seems Antoine wants to state that unless stated $i$ as imaginary explicitly, it has to be taken... Might be that is the cause? – jiten Jul 12 '22 at 07:33
  • @Zerox The answer is stated as wrong? – jiten Jul 12 '22 at 07:39

1 Answers1

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It sounds like you just entered into Wolfram Alpha without telling it to work in quaternions. It usually understands you mean the imaginary unit when you type i, but for j and k it usually treats them as variables not unit quaternions.

If you tell it you are in quaternions by putting an explicit call to that function, then you can get the objects you intended.

https://www.wolframalpha.com/input?i=Quaternion%5B1%2C-2%2C1%2C0%5D*Quaternion%5B0%2C1%2C0%2C-4%5D

Once you have the factors as quaternions through entering them within Quaternion[...] then you can multiply them.

AHusain
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