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The following matrix is given: $A=\begin{pmatrix} 2 & 1 & -2 \\ 0 & 1 & 0 \\ 0 & 3 & 4 \end{pmatrix} $

a) Determine kernel of matrix A

I did this, but I always end up with solution: $ker\left(A\right)=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ Can kernel be zero vector, is this solution possible?

b) Solve equation system $A\vec{x}=\vec{x} $ for $\vec{x} \neq 0$. Here I am getting confused with setting this system. I got:

$2x+y-2z=x$

$y=y$

$3y+4z=z$

Which doesnt look like correct solution. Can someone help me with setting this system?

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There is no problem with your kernel. It means that the application associated to the matrix is injective.

Your system is correct; it leads to $x=-3y=3z$ wich means that the set of solutions $\mathcal{S}=\text{Vect}\left\{\begin{pmatrix}-3\\1\\-1\end{pmatrix}\right\}$

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a) You are correct: the kernel consists of the zero vector only. Note that this vector is always in the kernel; in this case the kernel is a trivial subspace.

b) Your system is correct, now you have to continue solving it. Note that $y=y$ is trivially true and then from $3y+4z=z$ you have $z=-y$. Plugging this into the first equation allows to solve $x$ in terms of $y$ as well. You will have an infinite number of solutions.

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