Linear algebra question about matrices

  • Thread starter LaraCroft
  • Start date
  • #1
14
0
How would one determine the values of k such that the following matrix is the augmented matrix of a system with infinitely many solutions:

[ (k + 2) -2 1 | 2 ]
[ (k + 3) (k+ 3) 2 | 2 ]
[ (k + 2) -2 (k -1 ) | -3 ]

Also, how would I get all values of k such that the matrix for the same to be an augmented matrix of a system with no solutions?

It confuses me on how to find all possible values of k!

Thank you!:smile:
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
961
How would one determine the values of k such that the following matrix is the augmented matrix of a system with infinitely many solutions:

[ (k + 2) -2 1 | 2 ]
[ (k + 3) (k+ 3) 2 | 2 ]
[ (k + 2) -2 (k -1 ) | -3 ]

Also, how would I get all values of k such that the matrix for the same to be an augmented matrix of a system with no solutions?

It confuses me on how to find all possible values of k!

Thank you!:smile:
Try to row reduce it!
Reducing it to upper triangular form, so that you have only 0s below the main diagonal, you wind up with 0 0 f(k) | g(k). There is not a unique solution if f(k)= 0. There are an infinite number of solutions if g(k)= 0 also. (If f(k)= 0 and g(k) is not 0, then there is no solution.)
 
Last edited by a moderator:
  • #3
I would have set the determinant equal to zero. Of course the determinate can be calculated via row reduction.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
961
The problem with that is that if the determinant is 0, then the matrix equation may have infinitely many solutions or NO solution.
the question here was specifically to determine k so the equation has infinitely many solutions. You need to include the right hand side to determine that.
 
  • #5
The problem with that is that if the determinant is 0, then the matrix equation may have infinitely many solutions or NO solution.
the question here was specifically to determine k so the equation has infinitely many solutions. You need to include the right hand side to determine that.
You can always substitute the roots obtained by the determinate back into the original system and then do row reduction. You can also put the system into upper triangular form while computing the determinate.
 
  • #6
14
0
Ok...

Firstly, thank you everyone for responding...

Secondly, I am still not getting how to determine all the values of K, so that the matrix (call it A) is the augmented matrix of a system with infinitely many solutions. How would I find all values of K?

I understand that I need to row echelon it...but I think the way I am doing it is wrong...since the first entry cannot be 1...I start by doing Row 2 minus Row 1...and I I continue...however I am confused on what to do next in order to find k!

Apparently I should also be able to determine all values of k so that the matrix has a system with no solutions...

Thanks again:smile:
 
  • #7
Ok...
I understand that I need to row echelon it...but I think the way I am doing it is wrong...since the first entry cannot be 1...I start by doing Row 2 minus Row 1...and I I continue...however I am confused on what to do next in order to find k!
Before you do your subtraction multiply rows one and three by (K+3) and row 2 by (k+1)
 

Related Threads on Linear algebra question about matrices

  • Last Post
Replies
1
Views
2K
Replies
6
Views
10K
Replies
5
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
635
Replies
7
Views
3K
Replies
6
Views
557
Replies
2
Views
2K
Top