# Matrix Problem

• Oct 31st 2009, 01:55 PM
sf1903
Matrix Problem
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!
• Oct 31st 2009, 02:57 PM
mr fantastic
Quote:

Originally Posted by sf1903
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.
• Oct 31st 2009, 02:59 PM
tonio
Quote:

Originally Posted by sf1903
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):

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

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
$\displaystyle 2nd\,\;row\;:\;\;-3y+z=6\Longrightarrow y=-2+\frac{1}{3}z$
$\displaystyle 1st\,\;row\;:\;\;z-2+\frac{1}{3}z-z=4 \Longrightarrow x=6+\frac{2}{3}z$

As you can see, we wrote x,y as functions of z. Now just write $\displaystyle z=t$ 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!

.