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 Bucephalus January 28th, 2012 06:08 PM

Trivial solution only

Hi

I reduced a matrix to this and the book says it has only the trivial solution. For variables $x_{1}, x_{2}, x_{3} and x_{4}$

$\begin{bmatrix}
1 & 0 & \frac{-14}{4} & \frac{5}{2} & 0 \\
0 & 1 & 3 & -2 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$

For variables $x_{1}, x_{2}, x_{3} and x_{4}$ I got:

$x_{1}= \frac{14}{4}x_{3} - \frac{5}{2}x_{4}$

$x_{2}= -3x_{3} + 2x_{4}$

Then I assign the free variables arbitrary values r and s, so

$x_{3}= r$

$x_{4}= s$

such that:

$x_{1}= \frac{14}{4}r - \frac{5}{4}s$ and

$x_{2}= -3r + 2s$

But this isn't what the book has. It just says that it has the trivial solution only, which I can't easily see by inspection of this solution I have. Are they just saying that $x_{3} and x_{4}$ are
equal to zero since their rows are all zero rows?

David.

 Denis January 28th, 2012 08:18 PM

Re: Trivial solution only

Quote:
 Originally Posted by Bucephalus I reduced a matrix to this .....
WHAT matrix did you reduce?

 Bucephalus January 28th, 2012 08:24 PM

Re: Trivial solution only

This one

\begin{bmatrix}
0 & 1 & 3 & -2 & 0\\
2 & 1 & -4 & 3 & 0\\
2 & 3 & 2 & -1 & 0\\
-4 & -3 & 5 & -4 & 0
\end{bmatrix}

Thanks.

 Bucephalus January 28th, 2012 08:41 PM

Re: Trivial solution only

Ok I'm teaching latex more than getting my problem solved.
Here it is again.

$\begin{bmatrix}
0 & 1 & 3 & -2 & 0\\
2 & 1 & -4 & 3 & 0\\
2 & 3 & 2 & -1 & 0\\
-4 & -3 & 5 & -4 & 0
\end{bmatrix}$

 Grayham1990 January 28th, 2012 08:44 PM

Re: Trivial solution only

What does the question actually ask?

I double checked and your reduction is right.

In your steps it looks like you have gone on to find the null space of the matrix.

Did it ask to find the determinant of the matrix? Cause that has only the trivial solution since the matrix is singular.

Shouldn't there also be a fifth variable x5 since is a 4x5 matrix?

 Denis January 28th, 2012 08:45 PM

Re: Trivial solution only

$\begin{bmatrix}
0 & 1 & 3 & -2 & 0\\
2 & 1 & -4 & 3 & 0\\
2 & 3 & 2 & -1 & 0\\
-4 & -3 & 5 & -4 & 0
\end{bmatrix}$

Top row: b + 3c - 2d = 0
Next 3 rows:
2a + b - 4c + 3d = 0
2a +3b + 2c - d =0
-4a-3b + 5c -4d = 0

OK?

 Bucephalus January 28th, 2012 08:51 PM

Re: Trivial solution only

Yeah four variables.

 Bucephalus January 28th, 2012 08:52 PM

Re: Trivial solution only

The question said "Solve the homogeneous system of linear equations by any method."

 Bucephalus January 28th, 2012 09:02 PM

Stop!!

My bad, I was looking at the wrong answer in that back of the book.
Terribly sorry. Having a bad day.
I still haven't got the right answer yet though. Let me investigate more
$x_{1} = 7s - 5t
x_{2} = -6s + 4t
x_{3} = 2s
x_{4} = 2t$

They obviously used s and t as the arbitrary variables.
Now I will go back and see if I can get mine to look like this.
Sorry, welcome to complete this with me if you like.

 Bucephalus January 28th, 2012 09:04 PM

Re: Trivial solution only

Yes that works out now.