Consider for every real number a the linear system of equations: x + ( a + 1 ) y + a 2 z = a 3 ( 1 − a ) x + ( 1 − 2 a ) y = a 3 x + ( a + 1 ) y + a z = a 2 1. Find the solution for a = 2;2. Find the values of a for which the system has no solution, infinitely many solutions, and a unique solution;3. Find the solution for a = − 1.

Question

Consider for every real number a the linear system of equations:                    x        +        (        a        +        1        )        y        +                  a          2                z                            =                  a          3                                    (        1        −        a        )        x        +        (        1        −        2        a        )        y                            =                  a          3                                    x        +        (        a        +        1        )        y        +        a        z                            =                  a          2                    1. Find the solution for   a  =  2;2. Find the values of a for which the system has no solution, infinitely many solutions, and a unique solution;3. Find the solution for   a  =  −  1.

atoandro8f04v · Accepted Answer

You can write      (                            1                          a          +          1                                      a                      2                                                1          −          a                          1          −          2          a                          0                                      1                          a          +          1                          a                      )        (                            x                                      y                                      z                      )    =      A    ⋅        (                            x                                      y                                      z                      )    =      (                                        a                      3                                                            a                      3                                                            a                      2                                )    .In general, if   det  A  ≠  0,   A is invertible and you have a unique solution. If you subtract the third equation from the first you obtain  B      (                            x                                      y                                      z                      )    =      (                            0                          0                          a          (          a          −          1          )                                      1          −          a                          1          −          2          a                          0                                      1                          a          +          1                          a                      )        (                            x                                      y                                      z                      )    =      (                                        a                      2                    (          a          −          1          )                                                  a                      3                                                            a                      2                                )  Here  det  B  =  a  (  a  −  1  )  {  (  1  −  a  )  (  1  +  a  )  −  1  +  2  a  }  =      a      2    (  a  −  1  )  (  2  −  a  )  ,which vanishes for   a  =  0  ,  1  ,  2. For   a  =  1,   z is undetermined and you can have an infinite number of solutions. In this case you find   x and   y by substituting   a  =  1. If   a  =  0 then we have   x  =  y  =  0 and   z is undetermined. The case   a  =  2 you already considered.

Consider for every real number a the linear system of equations: <mtable columnalign="right lef

Answered question

Answer & Explanation

New Questions in Algebra I