Linear transformation from $R^3$ to $R^4$ How would I solve this question?
Let $T : R^3\rightarrow R^4$ be a linear map, if it is known that $T(2,3,1) =(2,7,6,−7)$, $T (0, 5, 2) = (−3, 14, 7, −21)$, and $T (−2, 1, 1) = (−3, 6, 2, −11)$, find the general formula for $L(x, y, z)$.
I feel like the equation would take the form $T[x,y,z] = [x,y,z,w]$ where the $ w$ can be replaced with some version of the other parts of the vector?
I copied and pasted this from where I saw it. I’m not sure if the ‘$L$’ was intended?
 A: Write the equations breaking them into basis terms 
Like $2T(e_1) +3T(e_2) +T(e_3) = v_1$
Now solve the matrix equation. You will get images of the standard basis elements. 
Now can you complete it? 
Note that, it even gives you whether the function is well defined or not. 
A: HintNote that $B:=\left\{\, \begin{bmatrix}   2 \\   3 \\1   \end{bmatrix}, \begin{bmatrix}   0 \\   5 \\2  \end{bmatrix} ,\begin{bmatrix}-2\\1\\1\end{bmatrix}\,\right\}$
express vector 
\begin{bmatrix}   x_1 \\   x_2 \\x_3 \end{bmatrix} as a linear combination of the basis vectors in $B$.Namely we find scalar $c_1,c_2,c_3$ satisfying
$\begin{bmatrix}   x_1 \\   x_2 \\x_3 \end{bmatrix}=c_1\begin{bmatrix}   2 \\   3\\1  \end{bmatrix}+c_2\begin{bmatrix}   0\\   5\\2  \end{bmatrix}+c_3\begin{bmatrix}-2\\1\\1\end{bmatrix}.$
This can be written as the matrix equation
$\begin{bmatrix}   x_1 \\   x_2 \\x_3 \end{bmatrix}=
P\begin{bmatrix}   c_1 \\   c_2  \\c_3\end{bmatrix}$
Where $P=\begin{bmatrix}   2 & 0&-2\\   3& 5&1\\1&2&1 \end{bmatrix}.$
Solve the equation and find the value of the $c_1,c_2,c_3$ . Then use linear transformation. You can easily find the formula for $T$.
