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

We determine the solution space of the system of linear equations

$I\,\,\,\,\,\,\,\,\,2x+3y\,\,\,\,\,\,\,\,\,\,\,\,\,-w=0$

$II\,\,\,\,\,\,4x+2y+2z+5w=0.$

We get

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

Thus we have an equivalent system in row echelon form.

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

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

is a solution.

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

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

Therefore

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

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

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

To the exercise