Find all values of $k$ for which the given augmented matrix corresponds to a consistent system

Question

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?

Answer 1

Well, I think what I failed to understand here is the algebra that allows me to deduce the unrestrictive or arbitrary nature of $k$, for my last system in my post converts to

$$\begin{align} x_1 + kx_2 &= -1 \\ (k-2)x_2 &= 0 \end{align}$$

We immediately see from the second equation that either $x_2 = 0$, which means that any value for $k$ will not pose any problems to our system, so it is consistent for all values of $k$ in this scenario; and if $k=2$, we also don't have any inconsistency.

Therefore, for all $k$ the system is consistent.

Answer 2

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.

Answer 3

When you take $k=2$ the rank of the matrix

A= \begin{bmatrix} 1&k\\0&k-2 \end{bmatrix}

becomes 1, which is same as that of the augmented matrix

\begin{bmatrix} 1&k&-1\\0&k-2&0 \end{bmatrix}

But for $k \neq 2 $ the rank of A will be 2 as the last row is non zero, which is still the same for the augmented matrix.

So $\forall k \in \mathbb{R} $ the system is consistent