Image of subspace under the matrix linear transformation Consider the linear transformation $\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ whose matrix in relation to the canonical base is:
$[T] = \begin{bmatrix}
        1 & 2 & -1 \\
        0 & 2 & 3 \\
        1 & -1 & 1 \\
        \end{bmatrix}$
What is the equation of the plane which is the image, through transformation $T$, of the subspace $x + y + 2z = 0$ of $\mathbb{R}^{3}$?
The solution is $4x + 7y + 9z = 0$. Can someone explain how to solve this?

$f(x)=(x+2y-z,2y+3z,x-y+z)$.
$x=-y-2z, (-y-2z,y,z)=y(-1,1,0)+z(-2,0,1)$, $(-1,-1,0), (-2,0,1)$ is the basis.
$f(-2,0,1)=(-3,3,-1)$. You have $4(-3)+7(3)+9(-1)=-12+21-9=0$.
$f(-1,1,0)=(1,2,-2)$ $4(1)+7(2)+9(-2)=4+14-18=0$.
This shows that $f(-2,0,1)$ and $f(-1,1,0)$ are in $4x+7y+9z=0$ since they generates $x+y+2z=0$ it implies that $f(x+2y+z=0)\subset 4x+7y+9z=0$. Now   show that $f(-2,0,1)$ and $f(-1,1,0)$  are linearly independent thus generate this plane.
