Let $T:\mathbb R^3 \to\mathbb R^2$ be defined by $T(a,b,c) = (a+b, 2a-c)$. Determine $T^{-1}(1,11)$

Question

Let $T:\mathbb R^3 \to \mathbb R^2$ be defined by $T(a,b,c) = (a+b, 2a-c)$. Determine $T^{-1}(1,11)$

I was solving some linear algebra problems and saw the above one. I remember that there is a theorem that states that $T$ is invertible iff $[T]_{\beta}^{\gamma}$ (here ${\beta},{\gamma}$ are the basises) is invertable. Moreover, I remember that $T: V \to W$ is invertable iff $\dim(V) = \dim(W)$. Hence, to my understending there must be no inverse for this $T$. However, it asks to find it so my understanding is faulty (I assume).

Please, can you clearify the situation?

Answer 1

You are correct that $T$ is not invertible, since that would require that $T$ is injective but that cannot occur since the dimension of $\mathbb R^2$ is less than that of $\mathbb R^3$. What $T^{-1}(1,11)$ likely means here is the preimage of $(1,11)$, i.e. the set of all $\mathbf x\in\mathbb R^3$ so that $T(\mathbf x)=(1,11)$. To do so, what you need to do is solve the matrix equation $$\begin{bmatrix}1&1&0\\2&0&-1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}1\\11\end{bmatrix}.$$

Answer 2

Nobody is asking you no work with the inverse of $T$ (which does not exist). The problem consists in computing$$T^{-1}(1,11)=\left\{(a,b,c)\in\mathbb R^3\,\middle|\,T(a,b,c)=(1,11)\right\}.$$It is a problem concerning the inverse image of something.

Answer 3

$T$ is a linear map between a space of dimension $3$ ($\mathbb R^3$) in a space of dimension $2$ ($\mathbb R^2$). Hence $T$ can't be invertible.

However, I think that you're asked to find the inverse image of the singleton $\{(1,11)\}$ under $T$. For that you don't need $T$ to be invertible. You just have to solve the equations:

$$\begin{cases} a+b &= 1\\ 2a-c &= 11 \end{cases}$$

which is equivalent to $x=(a,b,c) \in H = \{(a, 1-a, 2a-11) \mid a \in \mathbb R\}$.

Conclusion: the inverse image of $\{(1,11)\}$ under $T$ is the affine space $H$.

Answer 4

Find all solutions $(a,b,c)$ which satisfy $a+b=1$ and $2a-c=11$. You will notice that there are two equations and three variables. Therefore you have a one parameter family of solutions. So, maybe assume that $a$ is free. Solve for $b,c$ in terms of $a$.