You are given the matrix form of T and can determin image and kernel from that.

A vector, , is in the image of T if and only if there exist a vector, such that

Of course, you can solve that equation by row-reducing the augmented matrix

The vector is in the image of T if and only if row reduction does not result in an "impossible row"- a row where the first three numbers are 0 and the fourth is not.

A vector is in the kernel of T if and only if

which you can determine by row-reducing the augmented matrix

For this particular problem the easy way to do it is to look at the determinant! If the determinant is non-zero, then the matrix is invertible and so maps one to one onto . Its image is all of and its kernel is {0}.