Linear mapping/Kernel/2152 3-27-1 2-1-43/Exercise/Solution

We have to solve the system of linear equations

{\begin{matrix}2x&+y&+5z&+2w&=&0\\3x&-2y&+7z&\,\,\,\,-w&=&0\\2x&\,\,\,\,-y&-4z&+3w&=&0\,\end{matrix}}.

We eliminate with the help of the first equation the variable ${}y$ . The resulting system is ( ${}II'=II+2I$ , ${}III'=III+I$ )

{\begin{matrix}2x&+y&+5z&+2w&=&0\\7x&\,\,\,\,\,\,\,\,&+17z&+3w&=&0\\4x&\,\,\,\,\,\,\,\,&+z&+5w&=&0\,\end{matrix}}.

We eliminate from ${}II'$ with ${}III'$ the variable ${}z$ , this yields ( ${}II'-17III'$ )

{\begin{matrix}2x&+y&+5z&+2w&=&0\\4x&\,\,\,\,\,\,\,\,&+z&+5w&=&0\\-61x&\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\,&-82w&=&0\,\end{matrix}}.

We can now solve the system, because ${}x\neq 0$ determines the other variables. Set ${}x=82$ . Then ${}w=-61$ . Then

{}z=-4x-5w=-4\cdot 82-5(-61)=-328+305=-23\,.

Finally we get

{}y=-2x-5z-2w=-2(82)-5(-23)-2(-61)=-164+115+122=73\,.

The solution set, which is the kernel, is therefore

{\left\{s{\begin{pmatrix}82\\73\\-23\\-61\end{pmatrix}}\mid s\in \mathbb {R} \right\}}.