Determine the values of $k$ so that the following linear system has unique, infinite and no solutions.

Question

Determine the values of $k$ so that the following linear system has a unique solution, infinite solutions and no solution.

$2x + (k + 1)y + 2z = 3$

$2x + 3y + kz = 3$

$3x + 3y − 3z = 3$

I have tried to use the determinant of the matrix to solve but have got stuck.

\begin{bmatrix} 2 & k+1 & 2 \\ 2 & 3 & k \\ 3 & 3 & 3 \end{bmatrix}

I have found the determinant to be $3(k^2-3k+2)$ which I'm pretty sure is correct.

I start getting confused with what to do with the determinant now though. I've read a few of the questions here but I just don't get it - solving equal to $0$ gives $k=1$ and $k=2$, but how do I interpret these values?

Answer 1

Some ideas:

Let $A$ be the coefficient matrix of your system, $A = A(k)$, $\mathbf{b}$ the vector of independent terms and $\mathbf{x} = (x,y,z)^T$ the vector of unknowns. Then, you have a linear system such as:

$$ A \, \mathbf{x} = \mathbf{b}, $$ with augmented matrix $(A\vert \mathbf{b})$. Then, the Rouché-Fröbenius theorem tells us that:

• If $n = \mathrm{rank}(A)$, where $n = 3$ in your case, then the solution is unique.
• Otherwise, there are infinite many solutions.

If $\det{A}(k) = 0$ for those valid values of $k$ (you showed them to be $k \neq 1 \wedge 2 $), then the solution is unique and, since $A$ is invertible, is given by $\mathbf{x} = A^{-1} \mathbf{b}$. The opposite is not always valid, since for $k = 1$, for example, the first and third rows are coincident and, then, the system has infinite many solutions, given by:

$$\begin{align} & 2x + 3y + z = 3 \\ & 3x + 3y − 3z = 3 \\ \end{align}$$ Put $x = x(z)$ and $y = y(z)$ and you will get the equations for a straight line. Let us know what happens for $k = 2$.

Cheers!

Answer 2

If the determinant is not zero then the matrix is invertible, so there is a unique solution to the system $Ax = b$ given by $x = A^{-1}b$. If the determinant is zero then the matrix is singular it means that its kernel is not the equal to $\{0\}$. There are now two cases, if $b = (3,3,3)$ is in the image of $A$, then there is at least one solution $x_0$ and $x_0 + v$ is a solution for every $v$ in the kernel, i.e. there are infinitely many solutions (take any $v\neq 0$ then $x_0 + \lambda v$ is a solution for every $\lambda \in \mathbb{R}$ ). If $b$ is not in the image of $A$ then there is no solution.