For which $\lambda$ do we have solutions

Question

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.

Answer 1

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:

1. If $\lambda=0$, infinitely many solutions.
2. If $\lambda=-2$, infinitely many solutions.
3. If $\lambda=2$, no solution.
4. If $\lambda$ is not $0$ nor $2$ nor $-2$, the system has unique solution.

Answer 2

This depends on the system. $A$ will always have at least one solution if this is a homogenous system due to the trivial solution.

If you want no solutions and it is not homogenous, you simply need to create a row of contradiction, that is some row you have all $[0 \ 0 \ 0 .... \ 0] = [1]$, which is impossible, it is easy to find such a $\lambda$ by row reducing and manually figuring out.

Infinite solutions are generally present if there are free variables associated with the system (that is the number of leading 1's is less than the number of columns) and there are no rows that lead to no solutions.