Determine such values of k that the equation system has infinitely
many solutions. How many parameters these solutions depend on.
$\begin{cases}x+2y-3z+t=1\\x+4y+3z+4t=-4\\x-4y-21z-8t=k\end{cases}$
$\begin{bmatrix}1&2&-3&1|&1\\1&4&3&4|&-4\\1&-4&-21&-8|&k\end{bmatrix}$…after transformations i got $\begin{bmatrix}1&2&-3&1|&1\\0&1&3&\frac{3}{2}|&-\frac{5}{2}\\0&0&1&0|&\frac{k-16}{-6}\end{bmatrix}$
I don't know what to do next to finish the task
Best Answer
If $a(x+2y-3z+t)+b(x+4y+3z+4t)=x-4y-21z-8t,$
then $a+b=1$ and $2a+4b=-4$, so $a=\color{blue}4$ and $b=\color{purple}{-3}$.
If you multiply the first equation by $\color{blue}4$ and the second equation by $\color{purple}{-3}$ and add them,
you will get the third equation when $k=\,?$