\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt


QUESTION

For each of the following matrices find the kernel and the image
of the corresponding linear transformation $\tau$.

$\left[\begin{array}{ccc}-1&-1&0\\0&-1&-4\\1&0&-4\end{array}\right]
\hspace{1cm}
\left[\begin{array}{ccc}2&-1&3\\-4&2&-6\\34&-17&51\end{array}\right]
\hspace{1cm}
\left[\begin{array}{cccc}1&2&3&4\\4&3&2&1\\5&5&5&5\\3&1&-1&-3\end{array}\right]$

\bigskip


ANSWER

The first matrix has rank 2 and nullity 1. the vector (4,-4,1) is
a basis for $\ker\tau$, so $\ker\tau$ is the line $x=-y=4z$.
Choosing two independent vectors not on this line, such as (1,0,0)
and (0,1,0), the image im$\tau$ has as a basis the images of these
two vectors. So im$\tau$ is spanned by (-1,0,1) and (-1,-1,0)
which generate the plane $x-y+z=0$.

The second matrix has rank 1 and nullity 2. The kernel is the
plane $2x-y+3z=0.$ The vector (1,0,0) is not on this plane, it
maps to (2,-4,34), so the image is the line
$x=-\frac{y}{2}=\frac{z}{17}$

The third matrix has rank 2 and nullity 2. Gaussian elimination
shows that the kernel is the plane specified by $w=y+2z$ and
$x=-2y-3z$. Two independent vectors not in this plane are
(1,0,0,0) and (0,1,0,0) which map to (1,4,5,3) and (2,3,5,1)
respectively. Gaussian elimination on these shows that they span
the plane specified by $w=-y+z$ and $x=-y-z$.



\end{document}