Determine the value of a if the system $$\begin{align} x_1 + 4x_2 − 3x_3 + 2x_4 &= 0\\ 2x_1 + 7x_2 − 4x_3 + 4x_4 &= 0\\ −x_1 + ax_2 + 5x_3 − 2x_4 &= 0\\ 3x_1 + 10x_2 − 5x_3 + (a² + 4a + 1)x_4 &= 0 \end{align}$$ has more then 1 solution
I thought that perhaps I have to work with the Jordan reduction or Gausselimination, but I'm not sure how to do that.
We know that a homogeneous linear system has more than one solution if and only if the rank of the matrix of the system is less than the number of unknowns. See Rank-nullity theorem.
One of the possibilities how to find rank is to use row operations. You could start like this: $$\begin{pmatrix} 1 & 4 & -3 & 2 \\ 2 & 7 & -4 & 4 \\ -1 & a & 5 & -2 \\ 3 &10 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 4 & -3 & 2 \\ 0 &-1 & 2 & 0 \\ 0 &a+4& 2 & 0 \\ 3 &10 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 4 & -3 & 2 \\ 0 & 1 & -2 & 0 \\ 0 &a+4& 2 & 0 \\ 3 &10 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 4 & -3 & 2 \\ 0 & 1 & -2 & 0 \\ 0 &a+5& 0 & 0 \\ 3 &10 & -5 & a^2+4a+1 \end{pmatrix}$$
If $a+5=0$, i.e. if $a=5$, then you have a matrix which does not have full rank. So in this case you have that there is more than one solution.
If $a+5\ne 0$, you can divide the third row by $(a+5)$
$$\begin{pmatrix} 1 & 4 & -3 & 2 \\ 0 & 1 & -2 & 0 \\ 0 &a+5& 0 & 0 \\ 3 &10 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 4 & -3 & 2 \\ 0 & 1 & -2 & 0 \\ 0 & 1 & 0 & 0 \\ 3 &10 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & -3 & 2 \\ 0 & 0 & -2 & 0 \\ 0 & 1 & 0 & 0 \\ 3 & 0 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & -3 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 3 & 0 & -5 & a^2+4a+1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 3 & 0 & 0 & a^2+4a+1 \end{pmatrix}$$
The last matrix is rather simple, so I'll leave the rest for you.
You can also check on WolframAlpha for which values of $a$ determinant of this matrix is zero. (A square $n\times n$ matrix has full rank if an only if $\det A\ne 0$. So the values for which the determinant is zero are precisely the values you are looking for. This would not work for non-square matrices -- in this case you had the same number of equations as the number of unknowns.)