So I am trying to figure out if I am doing this problem correctly, I need to do the following Determine all values of scalars p;q and r for which given linear systems is consistent
$$\begin{Bmatrix} 4 & 6 & p \\ -2 & -1 & q \\ \end{Bmatrix}$$
so far I have it converted as follows ~
$$\begin{Bmatrix}
1 & 6/4 & p1/4 \\
-2 & -1 & q \\
\end{Bmatrix}$$
then ~ $$\begin{Bmatrix} 1 & 6/4 & p1/4 \\ 0 & 2 & qp1/2 \\ \end{Bmatrix}$$
then $$\begin{Bmatrix} 1 & 6/4 & p1/4 \\ 0 & 1 & qp1/4 \\ \end{Bmatrix}$$
then $$\begin{Bmatrix} 1 & 0 & -(p5/4) \\ 0 & 1 & qp1/4 \\ \end{Bmatrix}$$
from here I feel like the way this table reduced down I did something wrong, but if it isn't then I believe these would then be consistent since it has a unique solution.
side note there is suppose to be no r involved in this question
Thank in advance.
The matrix $$ \begin{bmatrix} 4 & 6 \\ -2 & 1 \end{bmatrix} $$ has nonzero determinant, meaning the matrix is nonsingular and so, its column space is all of $\mathbb{R}^2$, so there is a linear combination of the columns (a solution to the linear system of equations) that gives rise to $\begin{bmatrix} p \\ q \end{bmatrix}$, regardless of the values of $p$ and $q$.
In order to see this through row reduction, the row reduction should actually look like so: $$\left[\begin{array}{rr|r} 4 & 6 & p \\ -2 & 1 & q \end{array}\right] \implies \left[\begin{array}{rr|r} 1 & 3/2 & p/4 \\ -2 & 1 & q \end{array}\right] \implies \left[\begin{array}{rr|r} 1 & 3/2 & p/4 \\ 0 & 4 & p/2 + q \end{array}\right] \implies \left[\begin{array}{rr|r} 1 & 3/2 & p/4 \\ 0 & 1 & p/8 + q/4 \end{array}\right] \implies \left[\begin{array}{rr|r} 1 & 0 & p/16 - 3q/8 \\ 0 & 1 & p/8+q/4 \end{array}\right] $$
Again, we see that, this agrees with what we said earlier: the matrix is nonsingular, and so invertible, so regardless of $p,q$, we can always solve: \begin{align*} \begin{bmatrix} 4 & 6 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} p \\ q \end{bmatrix} \\ \begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} 4 & 6 \\ -2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} p \\ q \end{bmatrix} \\ &= \frac{1}{16}\begin{bmatrix} 1 & -6 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} \end{align*}