Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system?

1 h 2

5 20 8

Ok so I got 2h cannot = - 20

Then divide by 2 and got h cannot equal = -10?

Is this correct?

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- Jan 22nd 2013, 02:55 PMOldspice1212h such that the matrix is augmented...
**Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system?**

1 h 2

5 20 8

Ok so I got 2h cannot = - 20

Then divide by 2 and got h cannot equal = -10?

Is this correct? - Jan 22nd 2013, 03:10 PMPlatoRe: h such that the matrix is augmented...
- Jan 22nd 2013, 03:21 PMOldspice1212Re: h such that the matrix is augmented...
I have to make it into a

1 0

0 1

Is that what you mean? - Jan 22nd 2013, 03:32 PMPlatoRe: h such that the matrix is augmented...
- Jan 22nd 2013, 03:34 PMOldspice1212Re: h such that the matrix is augmented...
Ooh haha I got -4, does it look correct now?

- Jan 22nd 2013, 03:38 PMPlatoRe: h such that the matrix is augmented...
- Jan 22nd 2013, 03:54 PMOldspice1212Re: h such that the matrix is augmented...
Its 4 since (20-5h)x2=-2

So it's a augmented matrix of a consistant linear system if h cannot = 4? :D - Jan 22nd 2013, 04:07 PMSorobanRe: h such that the matrix is augmented...
Hello, Oldspice1212!

Quote:

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

. . $\displaystyle \left[\begin{array}{cc|c} 1&h&2 \\ 5&20&8 \end{array}\right]$

The system is consistent if its determinant isequal to zero.*not*

If $\displaystyle D = 0$, we have: .$\displaystyle \begin{vmatrix}1&h \\ 5&20\end{vmatrix} \:=\:0 $

. . Hence: .$\displaystyle 20 - 5h \:=\:0 \quad\Rightarrow\quad h \:=\:4$

$\displaystyle \text{The system is consistent for all }h \ne 4.$

- Jan 22nd 2013, 04:09 PMOldspice1212Re: h such that the matrix is augmented...
^ yup I got it, thanks guys :D