Show that if a square matrix satisfies A²-3A+I=0, it is nonsingular and its inverse is equal to A^(-1)=3I-A.

Question

chillout · Answer

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vishweshshrimali5 · Answer

umm... okay let me give you a hint..

$A^2 - 3A + I = 0$

$\implies A^2 - 3A = -I$

$\implies A(A - 3I) = -I$

$\implies A^{-1} A (A-3I) = A^{-1} (-I)$

vishweshshrimali5 · Answer

So can you find $A^{-1}$ from here ?

chillout · Answer

Yeah, I was just assuming a 2x2 matrix and doing all the dirty work. You put it plain simple. Thanks!

vishweshshrimali5 · Answer

:) Glad to be of help

chillout · Answer

My browser is a bit messed up. One moment while I restart it.

chillout · Answer

$A^{-1}\cdot A=I \I(A-3I)=-A^{-1}\rightarrow  (A-3I)=-A^{-1} \\rightarrow A-3I=-A^{-1},\ \  A^{-1}=3I-A$

vishweshshrimali5 · Answer

Great! Also, for an inverse to exist, the matrix must be non - singular so that proves it

chillout · Answer

Thanks. Just remembering some linear algebra and I'm so bad with this xD