Let A and B be the matrices
$$ A= \begin{pmatrix} 1 & 1 & α \\ 1 & 0 & 0 \\ 0 & 2 & β \\ \end{pmatrix} ,B = \begin{pmatrix} 1 & 1 & β \\ α & 0 & 0 \\ 0 & 3 & α \\ \end{pmatrix} $$
I've managed to reduce both matrices to its echelon form:
$$
A = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & α \\
0 & 0 & β-2α \\
\end{pmatrix}
$$
and
$$ B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & β \\ 0 & 0 & α-3β \\ \end{pmatrix} $$
However, I can't seem to form a linear system of equations with α and β separate from each other so I can find values for them.
Any kind of help is appreciated!! Thank you
Guide:
To make sure that the nullity only consist of the zero vector, make sure that $A$ is nonsingular.
To make sure that the column space of $B$ is $\mathbb{R}^3$, also make sure that $B$ is nonsingular.
That is we require their RREF to be nonsingular, or equivalently, every column has a pivot element.
Hint: