The question is: A is a square matrix where $3A^9- 7A^4 + 4A = I$. Prove that A is invertible by finding A^-1.
I have looked at other similar questions on this site: 1. Here 2. and Here
But they use definitions and properties that I haven't been taught. Supposedly it can be proven using the definitions we already have gotten.
We have learned nothing of eigenvalues yet, thought we have learned a little about determinants.
I have a rather dull solution but I don't think it is correct:
$3A^9- 7A^4 + 4A = I$
$3A^9- 7A^4 + 4A = A* A^{-1}$
$A^{-1}(3A^9- 7A^4 + 4A) = A^{-1}*A* A^{-1}$
$3A^8- 7A^3 + 4*I = A^{-1}$
From here it can be seen that if we multiply $A^{-1}$ by A we get the original definition given to us equivalent to I. Q.E.D.
I'm sure there is a better solution than this since it feels that this solution is not good since it works off of the assumption that A is invertible. A*A^-1 = I.
We haven't learned anything about Hermitian matrices so it would be helpful to try to solve this using basic inverse matrix definitions.
Woah. You're definitely overthinking here.
$M_n(R)$ is a ring for any other ring $R$, so we have the distributive property. (You can show this manually). Therefore, we have:
$$3A^9 -7A^4 + 4A = A(3A^8 - 7A^3 + 4) = I$$
And so...