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|
Best Answer
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 $k\not=3$