# Linear mapping/Kernel/2 3 0 -1/4 2 2 5/Exercise/Solution

We determine the solution space of the system of linear equations

${\displaystyle I\,\,\,\,\,\,\,\,\,2x+3y\,\,\,\,\,\,\,\,\,\,\,\,\,-w=0}$
${\displaystyle II\,\,\,\,\,\,4x+2y+2z+5w=0.}$

We get

${\displaystyle III=II-2\cdot I\,\,\,\,\,\,-4y+2z+7w=0.}$

Thus we have an equivalent system in row echelon form.

We choose first ${\displaystyle {}w=0}$ and ${\displaystyle {}z=2}$. Then ${\displaystyle {}y=1}$ according to III and according to I we get ${\displaystyle {}x=-{\frac {3}{2}}}$. Hence

${\displaystyle {\begin{pmatrix}-{\frac {3}{2}}\\1\\2\\0\end{pmatrix}}}$

is a solution.

We choose now ${\displaystyle {}w=1}$ and ${\displaystyle {}z=0}$. Then ${\displaystyle {}y={\frac {7}{4}}}$ according to III and according to I we get

${\displaystyle {}x=-{\frac {17}{8}}\,.}$

Therefore

${\displaystyle {\begin{pmatrix}-{\frac {17}{8}}\\{\frac {7}{4}}\\0\\1\end{pmatrix}}}$

is another solution, which is linearly independent of the first solution. Since the rank of the matrix is ${\displaystyle {}2}$, the kernel is two-dimensional, hence the kernel equals

${\displaystyle {\left\{a{\begin{pmatrix}-{\frac {3}{2}}\\1\\2\\0\end{pmatrix}}+b{\begin{pmatrix}-{\frac {17}{8}}\\{\frac {7}{4}}\\0\\1\end{pmatrix}}\mid a,b\in \mathbb {R} \right\}}.}$