- #1

mr_coffee

- 1,629

- 1

|7 -7|5 |

|21 h|15|

What am I trying to make h be? I row reduced to get

|7 -7 |5|

|0 21+h |0|

So if i let h = -21 will this make it have infin. solutions? because then i would have 0 0 0?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mr_coffee
- Start date

- #1

mr_coffee

- 1,629

- 1

|7 -7|5 |

|21 h|15|

What am I trying to make h be? I row reduced to get

|7 -7 |5|

|0 21+h |0|

So if i let h = -21 will this make it have infin. solutions? because then i would have 0 0 0?

- #2

TD

Homework Helper

- 1,022

- 0

- #3

mr_coffee

- 1,629

- 1

I'm supppose to find a value of k that will make it have no solutions.

I have:

|1 1 4 2|

|1 2 -4 3|

|6 13 k 20|

-6R2 + R3 -> R3

|1 1 4 2|

|1 2 -4 3|

|0 1 24+k 2|

now if i let k = -24 will this make it have no solutions? because 0 cannot equal 2 right?

- #4

TD

Homework Helper

- 1,022

- 0

You could also compute the determinant of the coëfficiënt matrix and let it equal 0, solve for k.

- #5

mr_coffee

- 1,629

- 1

I ended up with

|1 1 4 2|

|2 3 0 5|

|0 0 -20-k 0|

and now I'm stuck, anyu ideas?

- #6

TD

Homework Helper

- 1,022

- 0

Check your work again or use the other method I gave.

- #7

fahd

- 40

- 0

reduce the matrix in row echelon form,and follow the rules to solve it..its simple

Share:

- Replies
- 32

- Views
- 712

- Replies
- 18

- Views
- 281

- Last Post

- Replies
- 2

- Views
- 206

- Last Post

- Replies
- 8

- Views
- 711

- Replies
- 7

- Views
- 653

- Last Post

- Replies
- 3

- Views
- 388

- Last Post

- Replies
- 9

- Views
- 200

- Last Post

- Replies
- 8

- Views
- 498

- Replies
- 6

- Views
- 225

- Replies
- 10

- Views
- 327