How to frame this set of linear equations? 2 x + 1 y + 2 z = A 0 x + 2 y + 2 z = A 1 x + 2 y + 1 z = AI assume this can be rewritten as a matrix? How can I check if a solution exists such that x, y, and z are nonnegative?

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How to frame this set of linear equations?  2  x  +  1  y  +  2  z  =  A  0  x  +  2  y  +  2  z  =  A  1  x  +  2  y  +  1  z  =  AI assume this can be rewritten as a matrix? How can I check if a solution exists such that x, y, and z are nonnegative?

Marisol Morton · Accepted Answer

Use an augmented coefficient matrix, and obtain row-echelon form (using elementary row operations), to see if a solution exists, and/or if the system is inconsistent. If inconsistent, then no solution exists.  2  x  +  1  y  +  2  z  =  A  0  x  +  2  y  +  2  z  =  A  1  x  +  2  y  +  1  z  =  A  M  =      (                            2                          1                          2                          A                                      0                          2                          2                          A                                      1                          2                          1                          A                      )  Subtract 1/2 * (row 1) from row 3Multiply row 3 by 2Swap row 2 with row 3Subtract 2/3 * (row 2) from row 3Multiply row 3 by 3Subtract 1/3 * (row 3) from row 1Subtract 1/3 * (row 2) from row 1Divide row 1 by 2Divide row 2 by 3Divide row 3 by 6  Result  :        (                            1                          0                          0                          A                  /          6                                      0                          1                          0                          A                  /          3                                      0                          0                          1                          A                  /          6                      )  If you row reduce carefully, (and you should attempt this so you can gauge your success in being able to do so), you should obtain the following:  x  =  A  /  6  ,    y  =  A  /  3  ,    z  =  A  /  6So for any given value of A, you will have a unique solution for      (                            x                                      y                                      z                      )    =      (                            A                  /          6                                      A                  /          3                                      A                  /          6                      )  So long as   A  ≥  0, the solution will be non-negative.

How to frame this set of linear equations? 2 x + 1 y + 2 z = A 0 x +

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