 Solution

For a linear map $\alpha$, its kernel of a linear map and image of a linear map areBy 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 isThe 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 $D\ne 0$, the matrix is invertible andIf $D=0$, 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 givesIn 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 isThese 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 $0<b<2$,Therefore 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.