How can i show that the matrix inverse exists for A^2 - 3A + I = 0
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assuming that A is invertible rearrange the equation to make I the subject: $\displaystyle I = 3A - A^2$ multiply by $\displaystyle A^{-1}$ $\displaystyle A^{-1} = 3I - A$
umm.. show that it exists.. not find what it is if it exists
Originally Posted by ah-bee How can i show that the matrix inverse exists for A^2 - 3A + I = 0 $\displaystyle A^2 - 3A + I = 0$ $\displaystyle A^2 - 3A = -I$ $\displaystyle det(A^2 - 3A) = det(-I) \neq 0$ $\displaystyle det(A) . det(A - 3I) = det(-I) \neq 0$ Thus $\displaystyle det(A) \neq 0$
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