If ad−bc= α, then what is A^−1, in terms of α?

Asked Jun 29 '21 at 18:55

Active Jun 29 '21 at 19:00

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If $ad−bc= \alpha$, then what is $A^{−1}$, in terms of $\alpha$? Note that the $\alpha$ and $a$ are seperate variables.

I tried with some numbers for matrix $A$. For example I used: \begin{bmatrix}1&2\\4&9\end{bmatrix} and I got $1$ as the determinant.

The inverse of $A$ I found was \begin{bmatrix}9&-2\\-4&1\end{bmatrix}

I still don't unterstand the relationship between $\alpha = 1$ and $A^{-1}$. Could someone help please explain this one?

edited Jun 29 '21 at 19:00

Theo Bendit

asked Jun 29 '21 at 18:55

Roger Fed

Is $A$ supposed to be a $2 \times 2$ matrix? Try some more examples; some with determinant not $1$. – GEdgar Jun 29 '21 at 18:59
A is a nxn matrix it doesn't necessarily have to be a 2 x 2. Thanks GEdgar – Roger Fed Jun 29 '21 at 19:02
Thanks for the link Randall I think that helps answer this question. – Roger Fed Jun 29 '21 at 19:05
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A nontrivial relation is $\det A^{-1}=\alpha^{-1}$. – janmarqz Jun 29 '21 at 19:08
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Thanks so much janmarqz! – Roger Fed Jun 29 '21 at 19:09

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