How they deduce that $\det A=1$ just from the first coefficient and minor

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
• 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. – The Chaz 2.0 Jan 22 '15 at 19:10
• just try please – Educ Jan 22 '15 at 19:13
• How is the $A$ matrix whose determinant you're speaking about related to the $M$ defined in the first displayed formula? – Henning Makholm Jan 22 '15 at 19:44
• its just typo M=A – Educ Jan 22 '15 at 19:45

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 $\det A = \frac{8}{9} \frac{9^2}{72} = 1$.
• is that hold for any $1\leq i \leq n$ $$\frac{1}{\det A} C_{ii} = A_{ii}$$ – Educ Jan 22 '15 at 20:25
• for example: $$\frac{1}{\det A} C_{ij} = A_{kl}$$ – Educ Jan 22 '15 at 20:49
• No, obviously this is only true for corresponding elements. In your equation, you must have $i=k$, and $j=l$. – Victor Liu Jan 22 '15 at 21:07
$DetA = 1$ 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 $detA = \pm 1$).