Show that if a square matrix A satisfies $\displaystyle A^2 - 3A + I = 0$ then $\displaystyle A^-1 = 3I - A$

I tried to solve from a series of equations, but it just went in circles.

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- Jan 16th 2010, 11:11 AMBungkaiMatrixes Arithmetic
Show that if a square matrix A satisfies $\displaystyle A^2 - 3A + I = 0$ then $\displaystyle A^-1 = 3I - A$

I tried to solve from a series of equations, but it just went in circles. - Jan 16th 2010, 11:28 AMDinkydoe
Given is: $\displaystyle A^2-3A+I = 0 \Longrightarrow I = -A^2+ 3A$

Now observe that $\displaystyle A(3I-A) = -A^2+3A = I $. Hence $\displaystyle A^{-1} = 3I-A$