Proof of the affine property of normal distribution for a landscape matrix

One should approach this through characteristic functions. Recall that $X$ is normal $N(\mu,\Sigma)$ if and only if, for every deterministic vector $t$ of size $N\times1$, $$ E(\mathrm e^{\mathrm it'X})=\mathrm e^{\mathrm it'\mu-t'\Sigma t/2}, $$ where $t'$ denotes the transpose of $t$. For every $(A,b)$ of compatible sizes, if $Y=AX+b$, one gets $$ E(\mathrm e^{\mathrm it'Y})=\mathrm e^{\mathrm it'b}E(\mathrm e^{\mathrm it'AX})=\mathrm e^{\mathrm it'b}E(\mathrm e^{\mathrm is'X}), $$ where $s'=t'A$. Since $s=(t'A)'=A't$, one sees that $$ E(\mathrm e^{\mathrm it'Y})=\mathrm e^{\mathrm it'b}\mathrm e^{\mathrm is'\mu-s'\Sigma s/2}=\mathrm e^{\mathrm it'b+\mathrm it'A\mu-t'A\Sigma A't/2}. $$ The RHS is the characteristic function of the normal distribution $N(b+A\mu,A\Sigma A')$, and this is enough to identify the distribution of $Y$ as such. Note that no inverse or pseudo-inverse is involved and that this applies to $Y$ of any dimension.