Problem:
Find all values of k such that the matrices
$A = \left[ \begin{array}{cc} 1 & k + 1\\ -1 & 3k -2\\ \end{array} \right]$ and $B = \left[ \begin{array}{cc} 1 & 2\\ 3& 4\\ \end{array} \right]$ are not row - equivalent.
What I know and my work:
Def: Two matrices are row - equivalent if one can be converted into the other using elementary row operations.
If they are able to be converted into each other then that means that they should have the same RRE form right? Therefore I should be trying to compare their RREs.
FOR B:
$B = \left[ \begin{array}{cc} 1 & 2\\ 3& 4\\ \end{array} \right]$
R2 = R2-3R1
$B = \left[ \begin{array}{cc} 1 & 2\\ 0& -2\\ \end{array} \right]$
R2 = -2R2
$B = \left[ \begin{array}{cc} 1 & 2\\ 0& 1\\ \end{array} \right]$
R1 = R1 - 2R2
$B = \left[ \begin{array}{cc} 1 & 0\\ 0& 1\\ \end{array} \right]$
FOR A:
$A = \left[ \begin{array}{cc} 1 & k + 1\\ -1 & 3k -2\\ \end{array} \right]$
R2 = R2 + R1
$A = \left[ \begin{array}{cc} 1 & k + 1\\ 0 & 4k -1\\ \end{array} \right]$
This gives me k + 1 = 0, and 4k-1 = 1
Where do I go on from here? Or is this way completely wrong?
Lemmas:
$1.)$ Invertible matrices are row equivalent to one another.
Proof: I think you got this one.
$2.)$ If an invertible matrix $C$ is row equivalent to an arbitrary matrix $D$, then $D$ is also invertible.
Proof: If $C$ ~ $D$, there exists elementary matrices $E_1, ..., E_n$ such that $$C E_1 ... E_n = D$$
Multiplying both sides by $E_n^{-1} ... E_1^{-1} C^{-1}$, we get that $$I = D (E_n^{-1} ... E_1^{-1} C^{-1}) $$ so $(E_n^{-1} ... E_1^{-1} C^{-1}) = D^{-1}$
Your matrix $B$ is invertible since you can reduce it to the identity. In order to guarantee $A$ and $B$ are not row equivalent, we need that $A$ is not invertible. By your calculation, this is the case if $4k- 1 = 0$ and so
$k= 1/4$