Find the basis for the kernel and the image, for a linear mapping

Question

Let $T : \Bbb{R}^3 → \Bbb{R}^3$ given by $$T(x, y, z) = (x + 2y − z, y + z, x + y − 2z).$$

Using Gauss-array and reducing the system of equations to row echelon form I got:

$\{(3,-1,1)\}$ is a basis for the kernel and $\{(1,0,1),(2,1,-1)\}$ is a basis for the image.

Answer 1

The matrix of $T$ is $$ \begin{bmatrix} 1 & 2 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & -2 \end{bmatrix} $$ and the Gaussian elimination gives the reduced row echelon form $$ \begin{bmatrix} 1 & 0 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} $$ Thus a basis for the image is $\{(1,0,1),(2,1,1)\}$ (the first two columns of the matrix of $T$) and a basis for the kernel is obtained from the equations \begin{cases} x_1=3x_3\\ x_2=-x_3 \end{cases} so we get $\{(3,-1,1)\}$