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What 3 by 3 matrix E?

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3 Answers

(x, y, z+x)
= (x, y, z) + (0, 0, x)
So, the matrix is
1 0 0....0 0 0....1 0 0
0 1 0 + 0 0 0 = 0 1 0
0 0 1....1 0 0....1 0 1
Basically, you have equation (x,y,z+x)=E(x,y,z) or (x,y,z)+(0,0,x)=E(x,y,z), which transforms into
 
(0,0,x)=(E-I)(x,y,z), where I is the unit matrix, I={(1,0,0);(0,1,0);(0,0,1)} Now denote E-I as U, then we have:
 
U11x+U12y+U13z=0
U21x+U22y+U23z=0
U31x+U32y+U33z=x
 
Since x, y, and x are arbitrary, the only way to satisfy those three equations is to set all Uij equal to zero, except U31, which shall be set equal to 1. So U={(0,0,0};(0,0,0);(1,0,0)}
 
Then E can be easily found by adding U and I, E=U+I