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Yikes, I was solving this question as if x,y were vectors. I was getting completely stuck until I reread the problem and found that X and Y are integers. Then everything clicked. Got the same answers as below
My approach was to find A^2, leaving you with one linear equation for each of the nine cells.
Most of them are repeats; for the zero cells, you get five:
0 + 0x + 0y = 0
You also have this equation three times, along the diagonals:
1 + x + y = 0
And finally this equation, for the top-middle cell:
2 + x = 0
Which yields the solution that others have said below. My question is, is there a more systematic way of solving this?
There is a nice trick here in decomposing A as the identity plus a nilpotent matrix (or degree or order 2).
Shouldn't there be a matrix on the right hand side of the equation?
Because I think if I add up matrices the result is also a matrix.
You can see the 0 on the right side of the equation as:
[0 0 0]
X = -2, Y = 1.
Giving 2 ecuations:
1 + x + y = 0; <<< 1
2 + x = 0; <<< 2
clearing x in 2:
x = -2; <<< 3
replacing 3 in 1:
1 - 2 + y = 0;
y = 2 -1;
y = 1;
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