Consider the function $T : \mathbb{R}^3 \to \mathbb{R}^3$ given by
$$T(a, b, c) = (\operatorname{Curl }\vec{F}_{(a,b,c)} )(1, 1, 1),$$
where $\vec{F}_{(a,b,c)}$ is the vector field given by $$\vec{F}_{(a,b,c)}(x, y, z) = (axy + bxz, bxy + cyz, cxz + ayz).$$
Is $T$ a linear transformation? If so, calculate its matrix.
I know that I need to take the curl of $\vec{F}_{(a,b,c)}$, and then plug in the point $(1, 1, 1)$ for $(x, y, z)$. But how would I determine if $T$ is a linear transformation? Thank you for your help!
The map $T : \mathbb{R}^3 \rightarrow\mathbb{R}^3$ is a linear transformation just means that for any real numbers $\lambda_1,\lambda_2 \in \mathbb{R}$, and any points $\vec{x} = (a,b,c), \vec{y}=(a',b',c') \in \mathbb{R}^3$, it satisfy $$ T(\lambda_1\vec{x}+\lambda_2\vec{y}) = \lambda_1T(\vec{x}) + \lambda_2T(\vec{y}) $$ So for your mapping above, this means (denote $\vec{z} =\lambda_1\vec{x}+\lambda_2\vec{y}$) $$ (\text{Curl}\, F_{\vec{z}})(1,1,1) = \lambda_1 (\text{Curl}\, F_{\vec{x}})(1,1,1) + \lambda_2 (\text{Curl}\, F_{\vec{y}})(1,1,1) $$
$\textbf{Edit} :$
By $\vec{F}_{(a,b,c)}(x, y, z) = (axy + bxz, bxy + cyz, cxz + ayz)$, we have $$(\text{Curl}\, F_{(a,b,c)}) = (az-cy,bx-cz,by-ax).$$ So \begin{align} (\text{Curl}\, F_{(a,b,c)})(1,1,1) &= (a-c,b-c,b-a)\\ (\text{Curl}\, F_{(a',b',c')})(1,1,1) &= (a'-c',b'-c',b'-a') \end{align} And for $$\vec{z} = \lambda_1\vec{x}+\lambda_2\vec{y} = \lambda_1(a,b,c) + \lambda_2(a',b',c') = (\lambda_1 a+\lambda_2 a', \lambda_1 b+\lambda_2 b', \lambda_1 c+\lambda_2 c'), $$ we have \begin{align} &(\text{Curl}\, F_{\vec{z}})(1,1,1) = \Big((\lambda_1 a+\lambda_2 a')-(\lambda_1 c+\lambda_2 c') \Big) \hat{i} + \Big((\lambda_1 b+\lambda_2 b')-(\lambda_1 c+\lambda_2 c') \Big) \hat{j} + \Big((\lambda_1 b+\lambda_2 b')-(\lambda_1 a+\lambda_2 a') \Big) \hat{k} \\ &= \Big(\lambda_1 (a-c)+\lambda_2 (a' - c') \Big) \hat{i} + \Big(\lambda_1 (b-c)- \lambda_2 (b'-c') \Big) \hat{j} + \Big(\lambda_1 (b-a)-\lambda_2 (b'- a') \Big) \hat{k} \\ &= \lambda_1 (a-c,b-c,b-a) + \lambda_2(a'-c',b'-c',b'-a') \end{align} So $$ (\text{Curl}\, F_{\vec{z}})(1,1,1) = \lambda_1 (\text{Curl}\, F_{\vec{x}})(1,1,1) + \lambda_2 (\text{Curl}\, F_{\vec{y}})(1,1,1) $$
In matrix representation, $[T_{ij}]$ will map any vector $(a,b,c)$ to $T(a,b,c)$ by $$ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \mapsto \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ T_{21} & T_{22} & T_{23} \\ T_{31} & T_{32} & T_{33} \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} $$ Now, we observe that $T$ maped basis $(1,0,0)$ to a vector which is the entries is same as the entries of first column of the matrix $[T_{ij}]$. That is $$ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \mapsto \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ T_{21} & T_{22} & T_{23} \\ T_{31} & T_{32} & T_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} T_{11} \\ T_{21} \\ T_{31} \end{bmatrix} $$ By doing this for $(0,1,0)$ and $(0,0,1)$, we immidiately know that to compute the matrix transformation for $T$ we just have to compute $T(e_i)$ with $e_i$ is the standard $i^{th}$-basis in $\mathbb{R}^3$, $$ T(e_i) = \sum_{j=1}^{3}T_{ji}e_{j}. $$ So in conclusion, by computing $T(1,0,0)$, $T(0,1,0)$ and $T(0,0,1)$, you'll have the first, second and the third column of $[T_{ij}]$ respectively. I'll leave the details to you to work on.