For what value of k does the following system have a unique solution So I guess this question states that we know we have a unique solution we just want to know what value of "k"  will allow this. Would I even have to take the determinant since I already know there is a unique solution? Or is the answer I get from the determinant the value of k?

For what value of $k$ does the  following system of equations
  \begin{align*}x-3y&=6\\x+3z&=-3\\2x+ky+(3-k)z&=1\end{align*}
  have a unique solution?

$$ \left[
    \begin{array}{ccc|c}
      1&-3&0&6\\
      1&0&3&-3\\
      2&k&3-k&1
    \end{array}
\right] $$
Also when I take the determinant of this matrix can I ignore the numbers to the right of the horizontal line. 
det = 1|3k| + 3|(3-k)-6|+(0)|k|
 A: You can also do row reduction:
$\begin{bmatrix}1&-3&0&6\\0&1&1&-3\\0&k+6&3-k&-11\\\end{bmatrix}$
In order to have a unique solution, the third row cannot be [0 0 0], therefore $k+6\not=0, 3-k\not=0$
$k\not=-6$ and $k\not=3$
A: You are on the right track.
For the system to have
a unique solution,
the determinant has to be non-zero.
Simplify your expression
and see what you get.
If the determinant is zero,
then there are either
no solutions
or an infinite number of solutions
depending on the
right side.
A: Since it is given that you have to find suitable k for the problem to have unique solution asserts that problem has already unique solution. i.e. rank of matrix = rank of augmented matrix = no. of unknowns=3.
i.e.
rank of matrix should be 3.means determinant should be non zero.rank of augmented matrix should be 3 as well and rank of augmented matrix is always greater than or equal to rank of matrix.rank of augmented matrix can't be 4.
so,eliminate all values of k which gives matrix of determinant zero.rest all are answer to your problem.
