For which value of $k$ does the system of equations not have a unique solution?

Question

The following is the system of equations:
$$\begin{cases} x-2y+2z=0\\ 2x+ky-z=0\\ x-y+3z=-5 \end{cases}$$

Answer 1

Written in matrix form, this system is:

$$\begin{bmatrix}1 & -2 & 2\\ 2 &k& -1 \\ 1 & -1 & 3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\-5\end{bmatrix}.$$

In general, the matrix equation $\mathbf{A}\cdot \mathbf{x} = \mathbf{b}$ has a solution if and only if $\det{\mathbf{A}} \neq 0$.

To solve this problem, you can therefore compute the determinant of the 3×3 matrix and set it equal to zero. The system will have a solution for all values of $k$ except for those that make the determinant zero.

Answer 2

Putting the system of linear equations into a matrix, the next step is to try and use elementary row operations to get it to rref (reduced row echelon form). Let's not worry about the constants in the right hand side of the system of linear equations for now.

$$ \begin{bmatrix} 1 & -2 & 2 \\ 2 & k & -1\\ 1 & -1 & 3 \end{bmatrix} $$

Subtracting twice the first row from the second row and subtracting the first row from the third row, we get:

$$ \begin{bmatrix} 1 & -2 & 2 \\ 0 & k+4 & -5\\ 0 & 1 & 1 \end{bmatrix} $$

If $k=-9$, the second row is a scalar multiple of the third row. This means that if we had kept track of the augmented coefficient matrix, the second and third row would either equal the same thing, and thus cancel each other out, resulting in a free variable and infinite solutions, or equal different things, and thus be contradictory, meaning there would be no solutions to the system.

To confirm, we could check to make sure the second row, represented as a column vector, is not linearly independent of the other two equations:

$$ \begin{pmatrix} 2 \\ -9 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} p + \begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix} q $$

$$ p+q=2 \\ -2p-q=-9 \\ -2p-2+p=-9 \\ p=7 \\ q=-5 $$ Now, we can see that the second equation is just 7 times the first equation minus 5 times the third equation.

As a third method, one could check the determinant and discover that it's $k+9$, so when $k=-9$, the determinant is zero, and the system will not have a unique solution.