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

Question

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

Answer 1

$DetA = 1$ 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 $detA = \pm 1$).

Answer 2

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 $.