For what values of k in this set of linear equations A x = b has no solutions, an infinite number of solutions and a unique solutions?I know I want to be using Gaussian Elimination here, I've augmented the matrix and I'm perfectly familiar with ERO's and back-solving for systems without unknown constants but this is new to me. 2 2 0 2 0 k 1 1 1 2 k 2 Would I try to be putting this into Row-Echelon form? I have an inkling by playing with it that k = − 1 for no solutions and k = 1 for an infinite number of solutions. I can't do the Gaussian steps properly with a k involved to produce some decent working though.Thank you in advance for any help, solutions or tips. :)

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For what values of   k in this set of linear equations   A  x  =  b has no solutions, an infinite number of solutions and a unique solutions?I know I want to be using Gaussian Elimination here, I&#039;ve augmented the matrix and I&#039;m perfectly familiar with ERO&#039;s and back-solving for systems without unknown constants but this is new to me.                    2                    2                    0                    2                            0                    k                    1                    1                            1                    2                    k                    2            Would I try to be putting this into Row-Echelon form? I have an inkling by playing with it that   k  =  −  1 for no solutions and   k  =  1 for an infinite number of solutions. I can&#039;t do the Gaussian steps properly with a   k involved to produce some decent working though.Thank you in advance for any help, solutions or tips. :)

Alisa Jacobs · Accepted Answer

HintIn order to have unique solutions, the determinant should be nonzero :  det  (  A  )  =  0  ⇔                              2                          2                          0                                              0                          k                          1                                      1                          2                          k                      =  0  ⇔  2  (      k    2    −  1  )  =  0  ⇔  k  =  +  −  1Now, by plugging   k  =  1 to our matrix and doing a Reduced Echelon Form Transformation:      (                            1                          0                          −          1                          0                                      0                          1                          1                          1                                      0                          0                          0                          0                      )  and by plugging   k  =  −  1, executing again a Reduced Echelon Form Transformation:      (                            1                          0                          1                          0                                      0                          1                          −          1                          0                                      0                          0                          0                          0                      )  Can you now derive conclusions for inconsistent, unique solution and infinite solutions?

Ayaan Barr · Accepted Answer

As first comment already said you can use determinantats to find which answerz will dire t to a determinant equals to zero, which means a matrix which no aolution or with infinite solution. Using determinant you fimd that 1 and -1 are k values for determinant to be equals 0. Now if you try to use those two values and reduce the matrix to find the solution then you will find that k=1 lead to infinite solutions, since it leads to a system of three variables and two rows. On the other hand with k=-1 you will find at some point of resuction that two rows show contradictory information, for example one auggesting -x1 + x2 = 1 and the other one suggesting that -x1+x2=-2. In some way its like there are three rows but just two variables, since you have a row that you dont need in order to have a system of two rows two variables.

For what values of k in this set of linear equations A x = b has no solutions,

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