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

determinantlinear algebramatricessystems of equations

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?

Best Answer

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!

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