Linear Algebra - single solution

Question

I have a question and I'm not sure about my solution.

For which values of $\lambda$ the system has a single solution, an infinite number of solutions, and no solution? If there are solutions find them.

$$\left\{ \begin{eqnarray} -2x_1 + x_2 + x_3 =& -\lambda \\ x_1 + 2\lambda x_2 + x_3 =& 1 \\ x_1 + x_2 + 2\lambda x_3 =& 1\end{eqnarray} \right.$$

my answer:

• infinite number of solutions: $\lambda \in \{-1,2\}$
• no solution: $\lambda = -1$
• single solution: did not make it

Answer 1

You have a linear system of equations with three equations and three unknowns. It will have

• One unique solution if your coefficient matrix has full rank (3)

Otherwise, if it is rank deficient it can have:

• no solution if the RHS is not in the column space
• otherwise, infinitely many solutions (as it has a non-empty kernel)

In your case to check if the matrix is full rank, compute its determinant. You should get $$-8\lambda^2 - 4\lambda + 4 = (\lambda+1)(\lambda-\frac{1}{2}),$$ i.e., for all $\lambda \neq -1, \frac{1}{2}$ it is full rank and has an exact solution.

For $\lambda = -1$, the RHS becomes $[1,1,1]^{\rm T}$, which is not in the column space of the matrix (all three columns are orthogonal to $[1,1,1]^{\rm T}$. Hence, there is no solution.

For $\lambda = \frac{1}{2}$, the particular solution is $\frac{1}{4}[2,1,1]^{\rm T}$ and the kernel is $[0,1,-1]^{\rm T}$ so the set of solutions is $\frac{1}{4}[2,1,1]^{\rm T} + t \cdot [0,1,-1]^{\rm T}\quad \forall t \in \mathbb{R}$.

Answer 2

For the single solution, substituting $\lambda = 1$ into the associated 3x4 matrix (A) and converting A into Reduced Row Echelon Form, we see that $x_1 = 5/8, x_2 = 1/8, x_3 = 1/8$. Also, if $\lambda = 2$, Row Reducing A leads to $x_1 = 1, x_2 = 0, x_3 = 0$; rather than "infinite solutions" as you seem to suggest.

From the latter example, and in the context of the other answer, you should notice something about $\lambda = 2$.