Determine the value of $k$ such that the matrix is the augmented matrix of a linear system with infinitely many solutions.
$\left(\begin{array}{cc|r} 8 & -4 & 5\\ 16 & k & 10\\ \end{array}\right)$
Well if I divide row 2 by 2, I get $\left(\begin{array}{cc|r} 8 & -4 & 5\\ 8 & k/2 & 5\\ \end{array}\right)$
So when $k = -8$ both of these equations are the same and the bottom row will be all zeroes if I use the row operation Row 2 minus Row 1.
That means I will be left with $8x - 4y = 5$ which has infinitely many solutions. So the answer is $k = -8$. Is that correct? Is my reasoning correct?