If A is a matrix such that $A^2 - A - I=0$ then show that $(A + cI)$ is always invertible when $c\ne2$ and $c \ne1 $. I have no idea where to begin with this
2026-04-14 03:32:32.1776137552
Matrix invertibility
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If $A^2 = aA + bI$, it is easy to see that the inverse is given by $$A^{-1} = \frac{A-aI}{b}$$ (You can verify this by computing $A\frac{A-aI}{b}$).
But we want to find an inverse for $A+cI.$ To do this, we want to write an equation of the form $$(A+cI)^2=\lambda_1(A+cI)+\lambda_2I$$ so that we can use the above result. By using the identity $A^2=A+I$ given in the question, we can directly write $(A+cI)^2$ in the desired form. I think you can do the rest :)