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I found solution of exercise that said show that A is rotation to do that we have to compute det A=1 but they found it directly

Is there any relationship between the first coefficient and minor to say directly without complete computing that $\det A=1$

  • how they deduce that det A=1 just from the first coeffcient and minor
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    $\begingroup$ This appears to be a special case that relies on some sort of symmetry in the cofactor expansions along the top ow or left column. But I'm not sure how to explain it concretely. $\endgroup$ – The Chaz 2.0 Jan 22 '15 at 19:10
  • $\begingroup$ just try please $\endgroup$ – Educ Jan 22 '15 at 19:13
  • $\begingroup$ How is the $A$ matrix whose determinant you're speaking about related to the $M$ defined in the first displayed formula? $\endgroup$ – Henning Makholm Jan 22 '15 at 19:44
  • $\begingroup$ its just typo M=A $\endgroup$ – Educ Jan 22 '15 at 19:45
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I think the problem is having you try forming the product $A^TA$ and finding that it equals the identity. Therefore, $A^T = A^{-1}$.

Now you can use the equation given, where $$ A^{-1} = \frac{1}{\det A} C^T = A^T $$ Consider the first element, so $$ \frac{1}{\det A} C_{11} = A_{11} = \frac{8}{9}$$ Now, $C_{11} = 72/9^2$, so $\det A = \frac{8}{9} \frac{9^2}{72} = 1 $.

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  • $\begingroup$ is that hold for any $1\leq i \leq n$ $$\frac{1}{\det A} C_{ii} = A_{ii} $$ $\endgroup$ – Educ Jan 22 '15 at 20:25
  • $\begingroup$ It should hold for any element, not just the diagonal. $\endgroup$ – Victor Liu Jan 22 '15 at 20:41
  • $\begingroup$ for example: $$\frac{1}{\det A} C_{ij} = A_{kl}$$ $\endgroup$ – Educ Jan 22 '15 at 20:49
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    $\begingroup$ No, obviously this is only true for corresponding elements. In your equation, you must have $i=k$, and $j=l$. $\endgroup$ – Victor Liu Jan 22 '15 at 21:07
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$DetA = 1$ is a necessary but not sufficient condition for a matrix being a rotation. To be a rotation $A^T = A^{-1}$ (and here we include reflections composed with rotations, ie $detA = \pm 1$).

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  • $\begingroup$ My question wans't about rotation but it was about how they deduce that det A=1 just from the first coeffcient and minor $\endgroup$ – Educ Jan 22 '15 at 19:00

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