$$ A := \begin{pmatrix} 5 & 0 & 0 & 0 \\ 0 & 2 & 2 & 1 \\ 0 & 0 & -4 & 0 \\ 0 & 2a & 0 & -2 \end{pmatrix} $$ Find $\mathcal{L}=(A,0), a \in \mathbb{R}$
I thought all I had to do is to pring it in step form. So I did this:
$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & \frac{1}{2} \\ 0 & 0 & 1 & 0 \\ 0 & a & 0 & -1 \end{pmatrix} \\ I:5 \\ II:2 \\ III:-4\\ IV:2 $$
$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & \frac{1}{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -a & -1-\frac{a}{2} \end{pmatrix}\\ IV-aII $$
$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & \frac{1}{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1-\frac{a}{2} \end{pmatrix}\\ IV+aIII $$
$$ x_4 \cdot (-1-\frac{a}{2}) = 0\\ x_4 = 0 \\ x_3 = 0 \\ x_2 + 1 * 0 + \frac{1}{2} * 0 = 0 \\ x_2 = 0 \\ x_1 = 0 $$
$\Rightarrow \mathcal{L}=\{\begin{pmatrix} 0\\0\\0\\0 \end{pmatrix}\}$
I don't believe this is true. But I thought I have only done legal changes. Can someone explain what I did wrong? I surely did something that wasn't allowed.