Linear operator from matrix to general presentaition Let  $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ be a linear operator defined by:
$T(0,1,1)=(2,-1,1), T(2,-1,0)=(1,1,0), T(-1,0,0)=(1,-1,1)$.
I need to find $T(x,y,z)$. 
First of, i have found $[T]_{A}^{B}$ where A is the origin vectors and B is the image vectors that A is sent to.
I computed it to be
$$[T]_A^B=\begin{pmatrix} 1 & 0 & 2 \\
2 & - 1& 2 \\
2 & -3 & 3 \end{pmatrix}$$
But now after I have that matrix I am kind of lost. Is that even the right way to start?
 A: There are lots of ways to solve this problem. Here is a relatively straightforward srategy.
We are given that $T:\Bbb R^3\to\Bbb R^3$ is the linear map satisfying
\begin{align*}
T(0,1,1)&=(2,-1,1) & T(2,-1,0) &=(1,1,0) & T(-1,0,0)&=(1,-1,1)
\end{align*}
We wish to find a formula for $T(\vec x)$. To do so, we begin by solving the system
\begin{array}{rclclcl}
(1,0,0) & = & a_{11} (0,1,1) & + & a_{12} (2,-1,0) & + & a_{13} (-1,0,0)\\
(0,1,0) & = & a_{21} (0,1,1) & + & a_{22} (2,-1,0) & + & a_{23} (-1,0,0)\\
(0,0,1) & = & a_{31} (0,1,1) & + & a_{32} (2,-1,0) & + & a_{33} (-1,0,0)
\end{array}
which is equivalent to solving $AB = I$ where
\begin{align*}
A &=
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
&
B
&=
\begin{bmatrix}
0 & 1 & 1 \\
2 & -1 & 0 \\
-1 & 0 & 0
\end{bmatrix}
&
I &=
\begin{bmatrix}
1 & 0 & 0\\ 0&1&0\\0&0&1
\end{bmatrix}
\end{align*}
But multiplying $AB=I$ on the left by $B^{-1}$ gives $A=B^{-1}$ and one computes
$$
B^{-1}=\begin{bmatrix} 0&0&-1\\0&-1&-2\\1&1&2\end{bmatrix}
$$
This gives 
\begin{align*}
(1,0,0) &= -1\cdot(-1,0,0) \\
(0,1,0) &= -1\cdot (2,-1,0)+(-2)\cdot (-1,0,0) \\
(0,0,1) &= 1\cdot (0,1,1) +1\cdot(2,-1,0)+2\cdot(-1,0,0)
\end{align*}
Finally, we may compute
\begin{align*}
T(x,y,z)
&= T(x\cdot(1,0,0)+y\cdot(0,1,0)+z\cdot(0,0,1)) \\
&= x\cdot T(1,0,0)+y\cdot T(0,1,0)+z\cdot T(0,0,1) \\
&= ?
\end{align*}
Can you finish the computation?
A: Hint: Notice that $(0,1,1)$, $(2,-1,0)$, and $(-1,0,0)$ form a linearly independent set. On top of that, $(2,-1,1)$, $(1,1,0)$, and $(1,-1,1)$ form a linearly independent set. That means you already know $T$ with respect to the first basis (first set of vectors I wrote), and you know that it maps to another linearly independent set.
