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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?

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See this meta discussion for a suggestion about how to ask these questions - you might consider posting your own argument as an answer to your question. –  Zev Chonoles Feb 9 '12 at 2:30

1 Answer 1

up vote 1 down vote accepted

Yup, your reasoning is exactly right! Though ideally, you should explain why $k=-8$ is the only value of $k$ for which there are infinitely many solutions, since the problem says "the".

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