Correct me if I am wrong. Find the value(s) of the constant k such that the system of linear equations { x + 2 y = 1 k 2 x − 2 k y = k + 2 has:1. No solution2. An infinite number of solutions3. Exactly one solutionAnswer:so the first step is to get row reduction form, which is:from [ 1 2 k 2 − 2 k ] ,to [ 1 2 0 − 2 k + 2 k 2 ]

Question

Correct me if I am wrong. Find the value(s) of the constant k such that the system of linear equations      {                            x          +          2          y          =          1                                                  k            2                    x          −          2          k          y          =          k          +          2                        has:1. No solution2. An infinite number of solutions3. Exactly one solutionAnswer:so the first step is to get row reduction form, which is:from       [                            1                          2                                                  k            2                                    −          2          k                      ]  ,to       [                            1                          2                                      0                          −          2          k          +          2                      k            2                                ]

Jayce Bates · Accepted Answer

You row reduction is wrong. We get                    1                    2                    2                            0                    −        2                  k          2                −        2        k                    −                  k          2                +        k        +        2            which is equivalent to                    1                    2                    2                            0                    2        k        (        k        +        1        )                    (        k        +        1        )        (        k        −        2        )            From here we see that there is no solution iff   k  =  0, an infinite number of solutions iff   k  =  −  1 and else there is exactly one solution.

Feinsn · Accepted Answer

Hint:1. No solution when:      1                  k                    2              =      2          −      2      k        ≠      1          k      +      2      it holds when   k  =  02. An infinite number of solutions when:      1                  k                    2              =      2          −      2      k        =      1          k      +      2      it holds when   k  =  −  13. Exactly one solution when:      1                  k                    2              ≠      2          −      2      k      and it hold when   k  ≠  −  1

Correct me if I am wrong. Find the value(s) of the constant k such that the system of linear equatio

Answered question

Answer & Explanation

New Questions in Algebra I