Determine $a,b\in\mathbb R$ such that for linear transformation $f:\mathbb R^3\rightarrow \mathbb R^3$ is valid: $(4,3,4)\in Im(f)$.

Question

Determine $a,b\in\mathbb R$ such that for linear transformation $f:\mathbb R^3\rightarrow \mathbb R^3$ given by matrix $ \begin{bmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 2b & 1 \\ \end{bmatrix}$ in canonical basis is valid: $(4,3,4)\in Im(f)$.

Attempt:

$Im(f)=span\left(\begin{bmatrix} a \\ 1 \\ 1 \\ \end{bmatrix},\begin{bmatrix} 1 \\ b \\ 2b \\ \end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}\right)=\left\{c_1\begin{bmatrix} a \\ 1 \\ 1 \\ \end{bmatrix}+c_2\begin{bmatrix} 1 \\ b \\ 2b \\ \end{bmatrix}+c_3\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} : c_1,c_2,c_3\in\mathbb R\right\}$

We need to check for which $a,b\in\mathbb R$ the following system has a unique solution:

$$c_1a+c_2+c_3=4$$ $$c_1+c_2b+c_3=3$$ $$c_1+2c_2b+c_3=4$$

Using Kronecker-Capelli's theorem, the system will have a unique solution if $rank(A)=rank(A^{*})=3$ where $A$ is a $3\times 3$ and $A^{*}$ are $4\times 3$ matrices of the system.

Reduced echelon form of $A$ is $\begin{bmatrix} 1 & 2b & 1 \\ 0 & b & 0 \\ 0 & 0 & 1-a \\ \end{bmatrix}$ and for $A^{*}$ is $\begin{bmatrix} 1 & 2b & 1 & 4 \\ 0 & b & 0 & 1 \\ 0 & 0 & 1-a & \frac{4b-2ab-1}{b} \\ \end{bmatrix}$

$rank(A)=rank(A^{*})=3$ if $a=2,b\neq 0$.

Question: Is this the only possible combination for $a$?

Answer 1

The determinant of $A$ equals $b(1-a)$. Hence, if $b\neq 0$ and $a\neq 1$, the system always has a unique solution. It is clear that it has no solution if $b=0$. Thus, you only have to check the case where $a = 1$.