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Old October 31st, 2009, 02:55 PM
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Cool Matrix Problem

The question is, Solve the equation!

matrix-problem-matris1.gif

I think I have solved the first part...
x = 8
y = -1
z = 3

But then they say that the answer has to be written as;

matrix-problem-matris2.gif

Where a,b,c,d,e,f,g,h have to be itegers.

I have absolutely no idea how to solve this problem. Please help!


What is...

a=?
b=?
c=?
d=?
e=?
f=?
g=?
h=?

Thanks!
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  #2  
Old October 31st, 2009, 03:57 PM
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Quote:
Originally Posted by sf1903 View Post
The question is, Solve the equation!

Attachment 13603

I think I have solved the first part...
x = 8
y = -1
z = 3

But then they say that the answer has to be written as;

Attachment 13604

Where a,b,c,d,e,f,g,h have to be itegers.

I have absolutely no idea how to solve this problem. Please help!


What is...

a=?
b=?
c=?
d=?
e=?
f=?
g=?
h=?

Thanks!
You have not correctly solved the system of linear equations. From the first equation:

x = 4 - y + z.

The second and third equatins therefore become -3y + z = 6.

Let z = t where t is a free parameter that can equal any real number. Now solve for x and y in terms of t. The a, b, c etc. is simply telling you that your answer will involve fractions. If you do as I have said your answer will be in the required form.
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  #3  
Old October 31st, 2009, 03:59 PM
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Default

Quote:
Originally Posted by sf1903 View Post
The question is, Solve the equation!

Attachment 13603

I think I have solved the first part...
x = 8
y = -1
z = 3



This is is only one of the solutions: if you write down your augmented matrix, you get after simplifying it (by Gauss's method, say):

\left(\begin{array}{cccc}1&1&\!\!\!\!-1&4\\0&\!\!\!\!-3&1&6\\0&\!\!\!\!-9&1&18\end{array}\right)


As the third row is clearly a scalar multiple of the 2nd one it will vanish in the next step, and the general solution is
2nd\,\;row\;:\;\;-3y+z=6\Longrightarrow y=-2+\frac{1}{3}z
1st\,\;row\;:\;\;z-2+\frac{1}{3}z-z=4 \Longrightarrow x=6+\frac{2}{3}z

As you can see, we wrote x,y as functions of z. Now just write z=t and we're done.
This solution means: for ANY choice of the parameter t, you plug in this choice above for z and thus for x,y and you'll get a solution.

Tonio[/color]


But then they say that the answer has to be written as;

Attachment 13604

Where a,b,c,d,e,f,g,h have to be itegers.

[color=red]I have absolutely no idea how to solve this problem. Please help!


What is...

a=?
b=?
c=?
d=?
e=?
f=?
g=?
h=?

Thanks!
.
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