I am very raw at proofs, this is only my first semester learning them and I am having trouble with this problem. How would I approach this ?
Show that if a square matrix $A$ satisfies $A^2 - 3A + I = O$, then $A^{-1} = 3I - A$.
I am very raw at proofs, this is only my first semester learning them and I am having trouble with this problem. How would I approach this ?
Show that if a square matrix $A$ satisfies $A^2 - 3A + I = O$, then $A^{-1} = 3I - A$.
Actually nevermind, I got it! I will write the answer below for future reference.
If $A^2 - 3A + I = O,$ then, assuming $A^{-1}$ exists, multiply both sides of the equation by it: $A^{-1}(A^2 - 3A + I) = A^{-1}O = O.$ Expand the brackets: $A - 3I + A^{-1} = O.$ Leave $A^{-1}$ on the left: $A^{-1} = 3I - A.$
Remember that a square matrix $B$ is the inverse of a square matrix $A$ if $AB = I$ (or $BA = I$; each one implies the other). Using the equation for $A$, can you show that $A(3I - A)$ or $(3I - A)A$ is equal to the identity matrix?
Solve the equation for $I$, then factor the other side.