I have some problems, helpSolve the following nonhomogeneous system of linear equations: x 1 + 2 x 2 + 3 x 3 + x 4 = 4 , 2 x 1 + 2 x 2 + 3 x 3 + x 4 = 5 , 3 x 1 + 3 x 2 + 4 x 3 + x 4 = 6 Please enter the specific solution first and then the basis in the space of solutions of the corresponding homogeneous system.Comment of the teacherExample. If x s = ( 0 1 0 1 ) is a specific solution and { ( 1 2 3 / 2 4 ) , ( 0 − 2 2 9 ) } is a basis of { x : A x = 0 } then please enter[0,1,0,1],[1,2,3/2,4],[0,-2,2,9]There is an automated system here, checking the solution.SolutionAugmented matrix: ( 1 2 3 1 4 2 2 3 1 5 3 3 4 1 6 ) I managed to transform the augmented matrix to the diagonal form: ( 1 0 0 0 1 0 1 0 − 1 − 3 0 0 1 1 3 ) So, the specific solution is the last row + 0 at the bottom: x s = ( 1 − 3 3 0 ) . I checked, and it is really a solution of the original problem.1-st equation: 1 ⋅ 1 + 2 ⋅ ( − 3 ) + 3 ⋅ 3 + 1 ⋅ 0 = 1 − 6 + 9 = 42-nd equation: 2 ⋅ 1 + 2 ⋅ ( − 3 ) + 3 ⋅ 3 + 1 ⋅ 0 = 2 − 6 + 9 = 53-rd equation: 3 ⋅ 1 + 3 ⋅ ( − 3 ) + 4 ⋅ 3 + 1 ⋅ 0 = 3 − 9 + 12 = 6But, why does the basis in the comment consist of 2 columns? I think, it must be only 1: ( 0 − 1 1 − 1 ) - the 4-th column of the diagonal matrix and the last element (-1) is an element of negated identity matrix.When I tried this solution: [1, -3, 3, 0],[0, -1, 1, -1] for the first time, the automated checker said: "incorrect". Then I wrote this question; but before posting it, I tried it again, and it said: "correct". Probably, I made a typo first timeHow can I check, that ( 0 − 1 1 − 1 ) is really the basis of the space of solutions, like I did, validating the specific solution?

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I have some problems, helpSolve the following nonhomogeneous system of linear equations:                              x                      1                          +        2                  x                      2                          +        3                  x                      3                          +                  x                      4                                              =        4        ,                            2                  x                      1                          +        2                  x                      2                          +        3                  x                      3                          +                  x                      4                                              =        5        ,                            3                  x                      1                          +        3                  x                      2                          +        4                  x                      3                          +                  x                      4                                              =        6            Please enter the specific solution first and then the basis in the space of solutions of the corresponding homogeneous system.Comment of the teacherExample. If       x    s    =      (                            0                                      1                                      0                                      1                      )   is a specific solution and       {          (                                    1                                                2                                                3                          /                        2                                                4                              )        ,          (                                    0                                                −            2                                                2                                                9                              )        }   is a basis of   {  x  :  A  x  =  0  } then please enter[0,1,0,1],[1,2,3/2,4],[0,-2,2,9]There is an automated system here, checking the solution.SolutionAugmented matrix:      (                            1                          2                          3                          1                          4                                      2                          2                          3                          1                          5                                      3                          3                          4                          1                          6                      )  I managed to transform the augmented matrix to the diagonal form:      (                            1                          0                          0                          0                          1                                      0                          1                          0                          −          1                          −          3                                      0                          0                          1                          1                          3                      )  So, the specific solution is the last row + 0 at the bottom:       x    s    =      (                            1                                      −          3                                      3                                      0                      )  . I checked, and it is really a solution of the original problem.1-st equation:  1  ⋅  1  +  2  ⋅  (  −  3  )  +  3  ⋅  3  +  1  ⋅  0  =  1  −  6  +  9  =  42-nd equation:  2  ⋅  1  +  2  ⋅  (  −  3  )  +  3  ⋅  3  +  1  ⋅  0  =  2  −  6  +  9  =  53-rd equation:  3  ⋅  1  +  3  ⋅  (  −  3  )  +  4  ⋅  3  +  1  ⋅  0  =  3  −  9  +  12  =  6But, why does the basis in the comment consist of 2 columns? I think, it must be only 1:       (                            0                                      −          1                                      1                                      −          1                      )   - the 4-th column of the diagonal matrix and the last element (-1) is an element of negated identity matrix.When I tried this solution: [1, -3, 3, 0],[0, -1, 1, -1] for the first time, the automated checker said: &quot;incorrect&quot;. Then I wrote this question; but before posting it, I tried it again, and it said: &quot;correct&quot;. Probably, I made a typo first timeHow can I check, that      (                            0                                      −          1                                      1                                      −          1                      )  is really the basis of the space of solutions, like I did, validating the specific solution?

Hadley Cunningham · Accepted Answer

I&#039;m not sure if the teacher&#039;s comment was for specifically this question, but you&#039;re right. The solution to this system lies on a line and not a plane as the teacher&#039;s comment suggests.To check that your answer is correct, substitute the general form back into the system. Here&#039;s an example with the first equation:      x    1    +  2      x    2    +  3      x    3    +      x    4    =  4    ⟹    1  +  2  (  −  3  +      x    4    )  +  3  (  3  −      x    4    )  +      x    4    =  4    ⟹    1  −  6  +  9  +  3      x    4    −  3      x    4    =  4    ⟹    4  =  4

I have some problems, help Solve the following nonhomogeneous system of linear equations: <mta

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