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For a linear map $\alpha :{\mathbb{R}}^m\to {\mathbb{R}}^n$, its kernel and image of a linear map areThe kernel is a subspace of the domain and the image a subspace of the codomain. Given the stated bases, define the matrix of a linear map byIts $j$th column is the coordinate vector of $\alpha (e_j)$; if $v={\sum }_j{v}_j{e}_j$, the output coordinates are $(\alpha v{)}_i={\sum }_j{A}_{ij}{v}_j$.

For the change of basis, write $e_j^{\prime }={\sum }_k{P}_{kj}{e}_k$ and $f_i^{\prime }={\sum }_{\ell }{Q}_{\ell i}{f}_{\ell }$. The matrices $P,Q$ are invertible. Input coordinates change by $v_{\mathrm{old}}=P{v}_{\mathrm{new}}$ and output coordinates by $w_{\mathrm{old}}=Q{w}_{\mathrm{new}}$, soThe domain and codomain basis changes need not be the same matrix.

For $\beta$, the two input vectors form a basis of ${\mathbb{R}}^2$, so its image is the span of $v=(1,2,3{)}^T$ and $w=(6,4,2{)}^T$. The vector $n=(1,-2,1{)}^T$ is orthogonal to both, since $n\cdot v=n\cdot w=0$. Every possible output therefore satisfies $n\cdot \beta x=0$, whereas the target itself is $n$ and $n\cdot n=6$. Hence there is no $x$ with the required output. This uses an image obstruction by an annihilating functional, rather than an inconsistent guessed input.

For $\gamma$, use the input basis $v_1=(1,2,0{)}^T$, $v_2=(0,1,1{)}^T$, $v_3=(0,1,0{)}^T$. Its determinant is $-1$, so it is genuinely a basis. Write $x=s{v}_1+t{v}_2+w{v}_3$. ThenSubstitution back into physical coordinates givesFor a direct check, its standard matrix is $\left(\begin{array}{ccc}1& 0& -2\\ 1& 1& 0\end{array}\right)$, which annihilates exactly that line.