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?

As an alternative method, though it is perhaps a little longer:
First, find a basis of your subspace. As your subspace is a plane in $\mathbb{R^3}$, we know it will have dimension 2, so we need to find two linearly independent vectors which satisfy $x+y+2z = 0$. Through experimentation, I found that:
$$<1,1,-1>,<2,4,-3>$$
is a basis of the subspace, because each vector satisfies the desired equation and the two vectors are linearly independent (though, perhaps you could find a different basis that is easier to work with).
We next apply the linear transformation to the basis vectors to find the image of the subspace. I receive:
$$<4, -1, -1>, <13,-1,-5>$$
Which we note are two linearly independent vectors, and thus they will span a plane in $\mathbb{R}^3$.
We then can compute the cross product of these two vectors to find a vector orthogonal to them, and this will be the normal vector of the plane in the image, which we can use to write the equation of the plane.
A computation of the cross product produces the normal vector: $<4,7,9>$, and then the normal vector can be used to write the equation of the plane as:
$$4x+7y+9z=0$$
