What 3 by 3 matrix E?

Question

What 3 by 3 matrix E multiplies (x, y, z) to give (x, y, z+x)?

Answer 1

(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

Answer 2

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 U_ij 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

Answer 3

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]]