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?

Question

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?

vard6vv · Accepted Answer

It may make it easier to see what happens if you leave the ratio in the first row of the &quot;constants column&quot; 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&#039;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 &quot;indeterminate&quot; 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 &quot;dependent&quot; 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 &quot;inconsistent&quot; 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.)

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

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