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

Brennen Fisher

Brennen Fisher

Answered question

2022-05-22

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?

Answer & Explanation

vard6vv

vard6vv

Beginner2022-05-23Added 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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