Find the possible value/s that the constant c cannot take for the vectors to form a basis in $R^3$

Question

I have spent quite a lot of time on the following question:

The vectors

$v_1 = -ci + 6j + 2k$

$v_2 = 3i - cj + 2k$

$v_3 = -i + 2j + ck$

form a basis in the 3-dimensional space $R^3$. Find the possible value/s that the constant c cannot take.

I have tried forming a matrix with the above vectors and get a zero row at the very end, but this lead me with wrong answers $c = -6$ and $c = 2/3$.

I have arrived at the following matrix when conducting $3R_3 - R_2$ on the original set of vectors.

\begin{bmatrix}-c&3&-1\\6&-c&2\\0&6+c&3c-2\end{bmatrix}

The answer for this question is $c= \pm 4$ and $c = 0$.

Answer 1

The matrix is $A=\begin{bmatrix} -c & 6 & 2 \\ 3 & -c & 2 \\ 1 & 2 & c \end{bmatrix}$.

The determinant is $\det A = -c(-c^2-4) -6 (3c-2)+2(6+c)$. Now factor and find values for $c$ for which the determinant is zero.

Answer 2

You have to solve the system $$c\lambda_1+6\lambda_2+\lambda_3=0$$ $$6\lambda_1-c\lambda_2+2\lambda_3=0$$ $$\lambda_1+2\lambda_2+c\lambda_3=0$$ for $\lambda_1,\lambda_2,\lambda_3$