• Support PF! Buy your school textbooks, materials and every day products Here!

Matrices trouble, making a,b,c no sol, a unique, and infin solutions

  • Thread starter mr_coffee
  • Start date
  • #1
1,629
1
Hello everyone. I'm having troubles understanding if i'm doing this right.
I have the matrix
[tex]\left( {\begin{array}{*{20}c} 2 & {-3} & {-3} & a \\ {-1} & 1 & 2 & b \\ 1 & {-3} & 0 & c \\ \end{array} } \right)[/tex]

I row reduced it to:
[tex]\left( {\begin{array}{*{20}c} 2 & {-3} & {-3} & a \\ 0 & {-1} & 1 & 2b+a \\ 0 & {0} & 0 & {-3b-2a+c} \\ \end{array} } \right)[/tex]

I'm suppose to find, In each case find if possible conditions on the numbers a, b, and c that the given syhstem has no solution, a unique solution, or infintnety many s9lutions. So would i let b = 0, a = 0 and c equal 0 to make it have an infinit many solutions. and then let a, b and c be any number so the expression: -3b-2a+c wil not equal 0, so 0 = 4, no solution. How would i find a unique solution? if the last row is already 0 = -3b-2a+c? thanks!
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
In order to have at least one solution, the last row cannot by all 0 and a non-zero final element. So in order to have solutions, you have to let [itex]-3b-2a+c = 0[/itex].

Can you take it from here?
 

Related Threads for: Matrices trouble, making a,b,c no sol, a unique, and infin solutions

  • Last Post
Replies
6
Views
5K
Replies
13
Views
2K
Replies
1
Views
1K
Replies
1
Views
3K
Replies
12
Views
2K
Replies
1
Views
6K
  • Last Post
Replies
0
Views
881
Top