[Community Question] Linear-algebra: Find linear operator for given kernel and image

One of our user asked:

FInd linear map $A: \Bbb{R^3} \rightarrow \Bbb{R^3}$ for given kernel and image. $$Ker(A)=L(\begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix})\space ; \space Im(A)=L(\begin{pmatrix} 1 \\ 0 \\ 1 \\ \end{pmatrix}) \\$$ I've been reading some explonations about this kind of a problem but I didn't understand anything about expanding kernel base to the dimmension of $\Bbb{R^3}$. But, according to this solution example , if I form matrix $A$ like $$\begin{bmatrix} 1 & a&b \\ 0 &c&d\\ 1 &e&f \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \\$$ and $$ \begin{bmatrix} 1 & a&b \\ 0 &c&d\\ 1 &e&f \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \\ $$ I don't get anything here for first equation, and for second I get $$\begin{bmatrix} 1+a+d \\ b+e\\ 1+c+f \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\\$$ In this case I can't get a matrix of linear transformation like they did in linked example above. If someone can help me with this, but with theese concrete vectors, or to correct this way of solving..