Prove jacobian matrix is identity of matrix of order 3

[CrimsnDragn]

If f(x,y,z) = xi + yj +zk, prove that Jacobian matrix Df(x,y,z) is the identity matrix of order 3.

Because the D operator is linear, D1f(x,y,z) = i, D2f(x,y,z) = k, D3f(x,y,z) = k

There is clearly a relationship between this and some sort of identity, but I'm not sure how to state it, and I don't understand the order of linear transformations. Could someone help me?

 

[CrimsnDragn]

*typo on D2f(x,y,z) = j

actually I was just rethinking about the problem. could Df(x,y,z) = ((1,0,0),(0,1,0),(0,0,1)), which becomes an identity matrix, and the order of 3 refers to 3x3 matrix?

 

[Dick]

*typo on D2f(x,y,z) = j

actually I was just rethinking about the problem. could Df(x,y,z) = ((1,0,0),(0,1,0),(0,0,1)), which becomes an identity matrix, and the order of 3 refers to 3x3 matrix?

Yes, exactly.

 

[CrimsnDragn]

awesome. thanks!