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

Jun 16, 2018

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

#### Explanation:

have a look:

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.