How do you find ${(A - 2 I)}^{2} X = 0$ $(A-2I)^2X=0$ given $A=((3, 1, -2), (-1, 0, 5), (-1, -1, 4))$ ?

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How do you find ${(A - 2 I)}^{2} X = 0$ $(A-2I)^2X=0$ given $A=((3, 1, -2), (-1, 0, 5), (-1, -1, 4))$ ?

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Answer 1 · 2018-06-16T15:25:10

I tried this but I am not sure about it...I'll ask for review anyway.

Explanation:

have a look:
enter image source here
One solution would be (the trivial one):
$x_1=0,x_2=0,x_3=0$

I think that the three equations aren't independent (the first one is the third multiplyed by $-1$ while the second is the third multiplied by $2$ ), representing, geometrically, 3 coincident (one on top of the other) planes.

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How do you find ${(A - 2 I)}^{2} X = 0$ $(A-2I)^2X=0$ given $A=((3, 1, -2), (-1, 0, 5), (-1, -1, 4))$ ?

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How do you find (A-2I)^2X=0(A−2I)2X=0(A-2I)^2X=0 given A=((3, 1, -2), (-1, 0, 5), (-1, -1, 4))A=((3, 1, -2), (-1, 0, 5), (-1, -1, 4))?

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How do you find ${(A - 2 I)}^{2} X = 0$ $(A-2I)^2X=0$ given $A=((3, 1, -2), (-1, 0, 5), (-1, -1, 4))$ ?