- #1

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|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?

- Thread starter mr_coffee
- Start date

- #1

- 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

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- #3

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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

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

- #5

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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

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

- #7

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

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