Given $A^2 = A - I$ find $A^{15}$

Question

We should find matrix A such that $A^2 = A - I$ find $A^{15}$. I solved this with observing the pattern with raising A to different powers up to 6 and I realized that $A^{15} = -I$. However I'm not sure if this is the correct method, or if the result is correct.

Answer 1

$$A^2 = A - I\implies A^3= A(A^2) = A(A-I)= A^2-A=A-I-A=-I$$

Thus $$A^{15} = (-I)^5 = -I$$

Answer 2

You can do long division of the polynomial $x^{15}$ by the polynomial $x^2-x+1$ and you'll obtains polynomials $q(x)$ and $r(x)$ such that $x^{15}=(x^2-x+1)q(x)+r(x)$. This translates to an identity $$A^{15}=(A^2-A+I)q(A)+r(A)=0q(A)+r(A)=r(A)$$

Answer 3

We can observe that $A^3 + I = (A^2 - A + I)(A + I)$, and thus $A^{15} + I = 0.$

Answer 4

You can just whether verify your observation is correct: $$ A^6=A^4(A-I)=A^5-A^4=A^3(A-I)-A^4=-A^3. $$ And so $$ A^{15}=A^6A^6A^3=-A^6A^6=-A^3A^3=A^3=A(A-I)=A^2-A=A-I-A=-I. $$