= Solution
For a <linear map> $\alpha$, its <kernel of a linear map> and <image of a linear map> are
$$
K=\{x\in\mathbb R^3:\alpha x=0\},\qquad I=\{\alpha x:x\in\mathbb R^3\}.
$$
By the definition of the image, $\alpha x=y$ has a solution exactly when $y\in I$. If $x_0$ is one solution, every other solution is $x_0+k$ for some $k\in K$: subtracting the two equations proves necessity, and linearity proves sufficiency. Thus a solvable equation has a unique solution exactly when $K=\{0\}$.
Write the columns of the given matrix as $c_1=(1,0,1)^T$, $c_2=(1,t,t)^T$ and $c_3=(t,-2b,0)^T$. Its <determinant> is
$$
\det\alpha=-t^2+2bt-2b=-D,\qquad D=t^2-2bt+2b.
$$
The columns $c_1,c_2$ are <linearly independent> for every real $t$: proportionality would require the proportionality factor to be one from the first coordinate, then simultaneously $t=0$ from the second and $t=1$ from the third. Thus the <rank> is always at least two.
If $D\ne0$, the matrix is invertible and
$$
\boxed{K=\{0\},\qquad I=\mathbb R^3.}
$$
If $D=0$, the rank is exactly two. A direct substitution verifies the nonzero kernel vector $k=(-t^2,t,t-1)^T$: its first and third row products vanish identically and its second is $D$. The <rank-nullity theorem> then gives
$$
\boxed{K=\operatorname{span}\{(-t^2,t,t-1)^T\}.}
$$
In that case the image is the plane spanned by $c_1,c_2$. Their <cross product> is $(-t,1-t,t)^T$, so an equivalent description is
$$
\boxed{I=\operatorname{span}\{(1,0,1)^T,(1,t,t)^T\}=\{y:-ty_1+(1-t)y_2+ty_3=0\}.}
$$
These formulas include the singular case $t=b=0$, giving kernel $\mathbb R(0,0,1)$ and image $y_2=0$; no exceptional case was lost by dividing by $t$.
Finally, when $0<b<2$,
$$
D=(t-b)^2+b(2-b)>0\quad\text{for every real }t.
$$
Therefore \b[the equation has a unique solution for every $t$ and every $y\in\mathbb R^3$], in particular for every $y\in I$. Outside this parameter range, the singular cases are precisely $t=b\pm\sqrt{b(b-2)}$ when that square root is real.
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