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Find all real numbers $x$ for which there is a linear map $T:\mathbb{R}^3\to \mathbb{R}^3$ such that $T(1,1,1)=(1,x,1)$ and $T(1,0,-1)=(1,0,1).$

I am not sure how to proceed with this problem. I know that if $\{v_1,v_2,v_3\}$ are the basis of $\mathbb{R}^3$ then there exists a unique linear transformation $T$ such that $T(v_1)=u_1,T(v_2)=u_2$ and $T(v_3)=u_3$ where $u_1,u_2$ and $u_3$ are vectors in $\mathbb{R}^3.$ Any hints/advice regarding this problem will be much appreicated.

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You are almost there. Add a vector $v$ to the set $\bigl\{(1,1,1),(1,0,-1)\bigr\}$ such that $\bigl\{(1,1,1),(1,0,-1),v\bigr\}$ is a basis of $\mathbb{R}^3$. Then, by the theorem that you mentioned, there is a linear map in the conditions that you describe (choose $f$ such that $f(v)=(0,0,0)$, for instance), for each real number $x$.

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