what value of K does the system have a unique solution $\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$
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

Without solving you the problem, you can ALWAYS proceed with the following way if you don't spot any easy factorisation.
Suppose that, in your matrix, you have two values $a_{i,j}$ and $a_{i+1,j}$ you want to reduce in that way. You can ALWAYS exploit the property of linearity, and proceed with the substitution $R_{i+1} = a_{i,j}R_{i+1} - a_{i+1,j}R_{i}$, providing that $a_{i,j} \neq 0$.
In your case, proceed with the substitution: $R_3 = (-1-2K)R_3 - (10-K)R_2$, providing (i.e. check at the end if the solution allows it) that $-1-2K \neq 0$.
