Let $\alpha$ be a linear map

$$\alpha: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} .$$

Define the kernel $K$ and image $I$ of $\alpha$.

Let $\mathbf{y} \in \mathbb{R}^{3}$. Show that the equation $\alpha \mathbf{x}=\mathbf{y}$ has a solution $\mathbf{x} \in \mathbb{R}^{3}$ if and only if $\mathbf{y} \in I$

Let $\alpha$ have the matrix

$$\left(\begin{array}{ccc} 1 & 1 & t \\ 0 & t & -2 b \\ 1 & t & 0 \end{array}\right)$$

with respect to the standard basis, where $b \in \mathbb{R}$ and $t$ is a real variable. Find $K$ and $I$ for $\alpha$. Hence, or by evaluating the determinant, show that if $0<b<2$ and $\mathbf{y} \in I$ then the equation $\alpha \mathbf{x}=\mathbf{y}$ has a unique solution $\mathbf{x} \in \mathbb{R}^{3}$ for all values of $t$.