choosing h and k such that this system: { <mtable columnalign="left left" rowspa

Brennen Fisher 2022-05-22 Answered
choosing hand k such that this system:
{ x 1 + h x 2 = 2 4 x 1 + 8 x 2 = k
Has (a) no solution, (b) a unique solution, and (c) many solutions.
First i made an augmented matrix, then performed row reduction:
[ 1 h 2 4 8 k ] [ 1 h 2 0 8 4 h k 8 ]
Continuing row reduction, i get:
[ 1 0 k 8 2 ( h 2 ) + k 4 0 1 k 8 8 4 h ]
how to go about solving the problem with the matrix i end up with?
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Answers (1)

vard6vv
Answered 2022-05-23 Author has 12 answers
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:
[ 1 0 16 h k 4 ( 2 h ) 0 1 k 8 4 ( 2 h ) ]
As pointed out by Christiaan Hattingh, there's trouble if   h   =   2   for any value of   k   . So the system of equations is consistent (one solution) if   h     2  . If   h   =   2   and   k   =   8   , we would have   0 0   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:
[ 1 2 2 4 8 8 ]     ,
in which the second row is a multiple of the first; thus we have a "dependent" system with the single line equation   x + 2 y   =   2  . The remaining case is the one for which   h   =   2   and   k     8   ,, giving us a matrix
[ 1 2 2 4 8 8 ]     ,
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.)
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