Consider the augmented matrix
\begin{bmatrix}1&k&-1\\4&8&-4\end{bmatrix}
I want to find all the values of $k$ for which the corresponding linear system is consistent. Upon dividing the second row by $\frac{1}{4}$ we have
\begin{bmatrix}1&k&-1\\1&2&-1\end{bmatrix}
Upon subtracting the second row from the first row, and then assigning the results to the second row, we have
\begin{bmatrix}1&k&-1\\0&k-2&0\end{bmatrix}
In order to avoid a contradiction, it follows that $k$ can only equal $2$. However, the solution to this problem says that every $k\in \mathbb{R}$ causes the system to be consistent. What is the problem?
Best Answer
k = 2 is the only value of k that might form an inconstancy. But, it this case, it does not.
Lets turn this from an augmented matrix back to a system of linear equations.
$x + ky = -1\\ 4x + 8y = -4$
If $k \ne 2$ we have intersecting lines. If $k = 2$ then we have 2 equations for the same line.
No inconsistencies here.