# 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 $\mathbb{R}^{3}$

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

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

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

$x_{2}$ 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 $\begin{bmatrix} -3 & 1 & 0 \end{bmatrix}^{T} ,\, \begin{bmatrix} 2 & 0 & 1 \end{bmatrix}^{T}$ 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

• Maybe say that the kernel is spanned by those two vectors, and it looks good, very good! – pjs36 Feb 25 '15 at 2:55
• Okay, I already knew the answer was correct (it's in the back of the book) I just wanted to make sure that the way I was finding the kernel and presenting the results was adequate enough to receive full credit. – Linear Algebra Student Feb 25 '15 at 3:03