Reducing the System of linear equations x + 2 y − 3 z = 4 3 x − y + 5 z = 2 4 x + y + ( k 2 − 14 ) z = k + 2 I started doing the matrix of the system: ( 1 2 − 3 4 3 − 1 5 2 4 1 k 2 − 14 k + 2 ) But I still don't know how can I reduce this

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Reducing the System of linear equations                    x        +        2        y        −        3        z                            =        4                            3        x        −        y        +        5        z                            =        2                            4        x        +        y        +        (                  k          2                −        14        )        z                            =        k        +        2            I started doing the matrix of the system:      (                            1                          2                          −          3                          4                                      3                          −          1                          5                          2                                      4                          1                                      k            2                    −          14                          k          +          2                      )  But I still don&#039;t know how can I reduce this

scipionhi · Accepted Answer

Let us name the   3 by   3 matrix of the coefficients as   A and the   3 by   1 matrix of the constants as   B. And the matrix that you have provided becomes augmentation of two, name it   C  =  (  [  A  |  B  ]  )The system is said to be consistent, if we can find at least one solution, otherwise we say it inconsistent. To get an idea of consistency, find the determinant of   A- If   d  e  t  (  A  )  ≠  0, you have a unique solution and is consistent.- If   d  e  t  (  A  )  =  0, it can have either no solution (which is inconsistent) or infinitely many solutions (which is consistent).To proceed further with   d  e  t  (  A  )  =  0 condition:- If   R  a  n  k  (  A  )  ≠  R  a  n  k  (  C  ), you have no solution.- If   R  a  n  k  (  A  )  ≠  R  a  n  k  (  C  ) (the size of   A), you have infinitely many solutions. (If   R  a  n  k  (  A  )  =  R  a  n  k  (  C  )  =  n, you have a unique solution)You need to reduce the matrix so that you can find the Rank easily (otherwise, in my opinion, reducing the matrices is not going to help). The transformations you need to perform are:-       R    2    =      R    2    −  3      R    1   , it would make   A  [  2  ]  [  1  ] as   0.-       R    2    =      R    2    −  3      R    1   , it would make   A  [  3  ]  [  1  ] as   0.Continuing so, you should reduce the given matrix to an Upper Triangular matrix (reducing to Upper Triangular form will help you to find the Rank of the matrix)

Makayla Boyd · Accepted Answer

Let our augmented matrix be  A  =      (                            1                          2                          −          3                          4                                      3                          −          1                          5                          2                                      4                          1                                      k            2                    −          14                          k          +          2                      )  as you pointed out. So, if we do  −  3      R    1    +      R    2    →      R    2       (  I  )  ,           −  4      R    1    +      R    3    →      R    3       (  I  I  )  ,               R    2    −      R    3    →      R    3       (  I  I  I  )then we get  A  →  B  =      (                            1                          2                          −          3                          4                                      0                          −          7                          14                          −          10                                      0                          0                          16          −                      k            2                                    −          k          +          4                      )  Now think about-   k  =  4 which makes your system consistent with infinitely many solutions,-   k  =  −  4  which makes your system inconsistent,

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