Proper way to determine kernel and show results I am working on a HW assignment for a linear algebra course and I want to make sure that the way I am determining the kernel of a matrix and presenting the results is correct.  Here is the problem and my solution:

For the following matrix, write the kernel as the span of a finite number or vectors.  Is the kernel a point, line, plane or all of $\mathbb{R}^{3}$

$$ \begin{bmatrix} 2 &amp; 6 &amp; -4 \\ -1 &amp; -3 &amp; 2 \end{bmatrix} $$

The first step is to put the matrix into row-echelon form

$$\begin{bmatrix} 2 &amp; 6 &amp; -4 \\ -1 &amp; -3 &amp; 2 \end{bmatrix} \quad \xrightarrow{R_{1} + 2 R_{2}} \quad \begin{bmatrix} 2 &amp; 6 &amp; -4 \\ 0 &amp; 0 &amp; 0 \end{bmatrix} $$

$x_{2}$ and $x_{3}$ are now free, so we need to write an equation for $x_{1}$ in terms of $x_{2}$ and $x_{3}$

$$ 2x_{1} + 6x_{2} - 4x_{3} = 0 \quad\to\quad 2x_{1} = -6x_{2} + 4_{x3} \quad\to\quad x_{1} = -3x_{2} + 2x_{3} $$

this now allows us to determine the kernel

$$ \vec{x} \,\,=\,\, \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} \,\,=\,\, \begin{bmatrix} -3x_{2} + 2x_{3} \\ x_{2} \\ x_{3} \end{bmatrix} \,\,=\,\, x_{2} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix} \,\,+ \,\, x_{3} \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} $$

the kernel is $\begin{bmatrix} -3 &amp; 1 &amp; 0 \end{bmatrix}^{T} ,\, \begin{bmatrix} 2 &amp; 0 &amp; 1 \end{bmatrix}^{T}$ and is a plane.

As mentioned before, this is a HW assignment so I am simply looking for advice and, if needed, guidance in the right direction.

Thank you in advance for any help
That's certainly correct, you need nothing more!
That's certainly correct, you need nothing more!