Determine the value of $a$ for which the solution space is 2-dimensional

Question

The augmented matrix looks like \begin{pmatrix} 1 & 4 & -3 & 2 & 0\\ 2 & 7 & -4 & 4 & 0\\ -1 & a & 5 & -2 & 0\\ 3 & 10 & -5 & a^2 + 4a + 1 & 0 \end{pmatrix} I have to determine the value of $a$ for which the solution space 2-dimensional is, but I'm not sure where to start.

Answer 1

If, using Gaussian elimination, we use that $1$ at the top left corner to turn to $0$ the remaining entries of the first column, you get$$\begin{bmatrix}1 & 4 & -3 & 2 \\ 0 & -1 & 2 & 0 \\ 0 & a+4 & 2 & 0 \\ 0 & -2 & 4 & a^2+4 a-5\end{bmatrix}.$$Next, if you multiply the second column by $-1$ you get$$\begin{bmatrix}1 & 4 & -3 & 2 \\ 0 & 1 & -2 & 0 \\ 0 & a+4 & 2 & 0 \\ 0 & -2 & 4 & a^2+4 a-5\end{bmatrix}.$$Then if you use that $1$ at the second column and second row to turn the remaing entries of the second column into $0$, you get$$\begin{bmatrix}1 & 0 & 5 & 2 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 2 (a+5) & 0 \\ 0 & 0 & 0 & a^2+4 a-5\end{bmatrix}.$$Now, if $a\neq-5$, then the dimension of the space of solutions will be at least $3$. On the other hand, if $a=-$ then the all the entries of two final lines will be equal to $0$. And then the dimension of the space of solutions will be equal to $2$.