I was asked to check if there is a linear transformation that lives up to the requirements. If so, I should come up with T(x,y,x). Otherwise, I should explain why not.
a) $T:R^3 \to R^2$, so that $T(0,2,-1)=(-1,3)$, $T(2,-2,-1)=(1,4)$ and $T(1,-4,1)=(2,1)$.
b) $T:R^3 \to R^2$, so that $T(1,1,-1)=(-1,3)$, $T(1,1,1)=(1,4)$ and $T(1,0,1)=(2,1)$.
The way I resolved it was as follows: a) I created the following matrices: $$\begin{matrix}a&b&c\\d&e&f\end{matrix}$$
multiplied by
$$\begin{matrix}0&2&1\\2&-2&4\\-1&-1&1\end{matrix}$$
equals
$$\begin{matrix}-1&1&2\\3&4&1\end{matrix}$$
Then I wrote down the resulting two sets of equations. First:
2b-c=-1
2a-2b-c=1
a-4b+c=2
and resolved this through a matrix.
Next:
2e-f=3
2d-2e-f=4
d-4e+1f=1
and tried to resolve this via a matrix, but there was a contradiction, so no solutions. I assumed that meant that the transformation is not linear.
I did exactly the same for b), but this time the resulting two systems of equations both came out with one solution each. I assumed that this means this is indeed a linear transformation.
Is the reasoning correct?
Thank you!
Actually, the transformation in the first case is also linear. To find the matrix we just have to set up the linear system of equations $$ \begin{bmatrix}a&b&c\\d&e&f\end{bmatrix} \begin{bmatrix}0&2&1\\2&-2&4\\-1&-1&1\end{bmatrix}= \begin{bmatrix}-1&1&2\\3&4&1\end{bmatrix}. $$ The square matrix of the system is invertible, hence, the solution is $$ \begin{bmatrix}a&b&c\\d&e&f\end{bmatrix}= \begin{bmatrix}-1&1&2\\3&4&1\end{bmatrix} \begin{bmatrix}0&2&1\\2&-2&4\\-1&-1&1\end{bmatrix}^{-1} $$ Similar system can be solved in the second case.