 # choosing h and k such that this system: { <mtable columnalign="left left" rowspa Brennen Fisher 2022-05-22 Answered
choosing $h$and $k$ such that this system:
$\left\{\begin{array}{l}{x}_{1}+h{x}_{2}=2\\ 4{x}_{1}+8{x}_{2}=k\end{array}$
Has (a) no solution, (b) a unique solution, and (c) many solutions.
First i made an augmented matrix, then performed row reduction:
$\left[\begin{array}{ccc}1& h& 2\\ 4& 8& k\end{array}\right]\sim \left[\begin{array}{ccc}1& h& 2\\ 0& 8-4h& k-8\end{array}\right]$
Continuing row reduction, i get:
$\sim \left[\begin{array}{ccc}1& 0& \frac{k-8}{2\left(h-2\right)}+\frac{k}{4}\\ 0& 1& \frac{k-8}{8-4h}\end{array}\right]$
how to go about solving the problem with the matrix i end up with?
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It may make it easier to see what happens if you leave the ratio in the first row of the "constants column" in your reduced matrix over a common denominator:
$\left[\begin{array}{ccc}1& 0& \frac{16-hk}{4\left(2-h\right)}\\ 0& 1& \frac{k-8}{4\left(2-h\right)}\end{array}\right]$
As pointed out by Christiaan Hattingh, there's trouble if for any value of . So the system of equations is consistent (one solution) if . If and , we would have in both entries of the constant column. This would be interpreted to mean that both variables take on "indeterminate" values. The original matrix looks like:

in which the second row is a multiple of the first; thus we have a "dependent" system with the single line equation . The remaining case is the one for which and , giving us a matrix

which is now an "inconsistent" system, one with no solutions. (In the row-reduced augmented matrix at the start of this post, the entries in the constants column become two ratios having a non-zero number divided by zero.)