For which $\lambda$ do we have solutions I'm trying to find for what values of $\lambda$ the following matrix has either no solutions, infinitely many or unique solutions. 
$$A=\begin{pmatrix} 1 & 1 & \lambda & 1 \\ 4 & \lambda ^2 & -8 & 4 \\ \lambda & 2 & 4  & \lambda + 1\end{pmatrix}$$
I know that in general a system of equations has no solutions if there is a leading $1$ in the last column, unique if there is a leading $1$ in every column except the last and infinitely many otherwise. I'm not sure how to apply this here though. 

Because of the way your question is phrased, I will assume that $A$ is the augmented matrix of a system of 3 equations on 3 variable
If we perform row reduction:
\begin{align}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 4 & \lambda ^2 & -8 & 4 \\ \lambda & 2 & 4  & \lambda + 1\end{bmatrix}
&\xrightarrow[R3-\lambda R1]{R2-4R1}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 0 & \lambda ^2 -4& -8-4\lambda & 0 \\ 0 & 2-\lambda & 4-\lambda^2  &   1\end{bmatrix}\\ \ \\
&=
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 0 & (\lambda -2)(\lambda+2)& -4(\lambda+2) & 0 \\ 0 & 2-\lambda & -(\lambda-2)(\lambda+2)  &   1\end{bmatrix}
.
\end{align}


*

*If $\lambda=2$, we get
$$
\begin{bmatrix} 1 & 1 & 2 & 1 \\ 0 & 0& -16 & 0 \\ 0 & 0 & 0  &   1\end{bmatrix}
$$
The fourth row now signifies the equation $0=1$, so the system has no solution. 

*If $\lambda=-2$, we get
$$
\begin{bmatrix} 1 & 1 & -2 & 1 \\ 0 & 0& 0 & 0 \\ 0 & 4 & 0  &   1\end{bmatrix}
\xrightarrow{}
\begin{bmatrix} 1 & 0 & -2 & 3/4  \\ 0 & 1 & 0  &   1/4\\ 0 & 0& 0 & 0\end{bmatrix}
$$
This says that $x_2=1/4$, and $x_1=2x_3+3/4$. As we are free to choose $x_3$, the system has infinitely many solutions. 

*If $\lambda$ is neither $2$ nor $-2$, we can divide the second row by $\lambda+2$ and the third one by $\lambda-2$ to get
\begin{align}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 0 & \lambda -2& -4 & 0 \\ 0 & -1& -\lambda+2  &   1\end{bmatrix}
&\xrightarrow[R2+(\lambda+2)R3]{R1+R3}
\begin{bmatrix} 1 & 0 & 2 & 2 \\ 0 & 0& -\lambda^2 & 0 \\ 0 & -1& -\lambda+2  &   1\end{bmatrix}\\ \ \\
&\xrightarrow[]{}
\begin{bmatrix} 1 & 0 & 2 & 2  \\ 0 & 1& \lambda-2  &   -1\\ 0 & 0& \lambda^2 & 0\end{bmatrix}.
\end{align}
If $\lambda\ne0$, then the system will have three leading ones and no inconsistency, so it has unique solution. 
If $\lambda=0$, then again we are free to choose $x_3$ and so the system has infinitely many solutions. 
In summary:


*

*If $\lambda=0$, infinitely many solutions. 

*If $\lambda=-2$, infinitely many solutions. 

*If $\lambda=2$, no solution. 

*If $\lambda$ is not $0$ nor $2$ nor $-2$, the system has unique solution. 

