Robert J. answered • 11/06/13

Certified High School AP Calculus and Physics Teacher

Robert J. answered • 11/06/13

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(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

Kirill Z. answered • 11/07/13

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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:

U_{11}x+U_{12}y+U_{13}z=0

U_{21}x+U_{22}y+U_{23}z=0

U_{31}x+U_{32}y+U_{33}z=x

Since x, y, and x are arbitrary, the only way to satisfy those three equations is to set all U_{ij} equal to zero, except U_{31}, 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

Andre W. answered • 11/06/13

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E must be the identity matrix with an added 1 in the 1st column, 3rd row:

E = [[1,0,0],[0,1,0][1,0,1]]

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