$\begin{cases}x_1 + kx_2 – x_3 = 2\\2x_1 – x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$
I've been trying echelon form where i took $R_2 = R_2 – 2R_1$ and $R_3 = R_3-R_1$
So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$
I've been trying echelon form where i took $R_2 = R_2 – 2R_1$ and $R_3 = R_3-R_1$
and reduced it
So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\0&-1-2K&K+2&1\\0&10-K&-5&-1\end{array}\right]$
But now I am not sure how i could remove $10-K$ with $-1-2K$ any help would be appreciated
Best Answer
If $k\neq -\frac12$ $$\left( \begin{matrix} 1 & k & -1 & | & 2\\ 2 & -1 & k & | & 5 \\ 1 & 10 & -6 & | & 1 \\ \end{matrix} \right) \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left( \begin{matrix} 1 & k & -1 & | & 2\\ 0 & -1-2k & k+2 & | & 1 \\ 0 & 10-k & -5 & | & -1 \\ \end{matrix} \right) \xrightarrow[\text{$(2k+1\neq0)$}]{R_3=R_3-(\frac{10-k}{2k+1})R_2} $$ $$\left( \begin{matrix} 1 & k & -1 & | & 2\\ 0 & -1-2k & k+2 & | & 1 \\ 0 & 0 & \frac{-k^2-2k+15}{2k+1} & | & \frac{-3k+9}{2k+1} \\ \end{matrix} \right) $$ Which has a unique solution $\iff \frac{-3k+9}{2k+1} \neq 0 \iff -3k+9\neq 0 \iff k\neq3$
If $k=-\frac12$ it is easy to show that the system has a unique solution.