I want to see if it is possible to find a basis in $ℝ^3$ such that $A$ can be written in a special form Let us consider the matrix:
$$A = \begin{bmatrix}
a & b & c \\
d & k & f \\
m & v & r \end{bmatrix}$$
I want to see if it is possible to find a basis in $ℝ^3$ such that $A$ can be written as follow:
$$B = \begin{bmatrix}
z & w & 1 \\
s & q & 1 \\
x & p & 1 \end{bmatrix}$$
 A: This is possible if and only if $A$ is not a scalar multiple of the identity matrix $I$.
If $A$ is a scalar multiple of $I$, it is only similar to itself. Hence no change of basis would turn $A$ into the desired form.
Suppose $A$ is not a scalar multiple of $I$. Then there exists some nonzero vector $w$ such that $Aw$ is not a scalar multiple of $w$ (why?). Hence $x=Aw-w$ and $w$ are linearly independent. Let $\{x,y,w\}$ be a basis of $\mathbb R^3$. Set $u=x-y$ and $v=y$. Then $u,v,w$ are linearly independent. Therefore $P=[u,v,w]$ is invertible.
Now, let $\mathbf1$ be the all-one vector and let also $B=P^{-1}AP$. Since $PB=AP$, if the last column of $B$ is $b$, we have $Pb=Aw=x+w=u+v+w=P\mathbf1$. Therefore $b=\mathbf1$.
A: Hint:Let's assume that under a certain basis $E=[\epsilon_1,\epsilon_2,\epsilon_3]$ in $\mathbb R^3$ there is a linear transformation $T$ whose matrix is
$$A = \begin{bmatrix}
a & b & c \\
d & k & f \\
m & v & r \end{bmatrix}$$
And what we need to do is to find a suitable basis $E'=[\epsilon'_1,\epsilon'_2,\epsilon'_3]$ in $\mathbb R^3$ under which $T$'s matrix is 
$$B = \begin{bmatrix}
z & w & 1 \\
s & q & 1 \\
x & p & 1 \end{bmatrix}$$
Then we have the following equation
$$B=M^{-1}AM$$
where
$$E'=EM$$
($M$ is called the transition matrix from $E$ to $E'$, and it's obvious that $M$ is non-singular.)
Then if you can find some matrix $B$ s.t. $B$ is similar to $A$, then you're done.
