Math problem (1 Viewer)

Loppan

Senior Member
Jul 13, 2002
2,528
#1
I have these equations, I need to figure out how to show how I got the answer that a could be 2 if x=2, y=0 and b=b ( I ran it in a computer program) I know I have to use Gauss elinimination but how??? :

{x-y+a*z=1, 2*x-y+z=-1, a*x+y-z=1};
 
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Loppan

Loppan

Senior Member
Jul 13, 2002
2,528
#3
hm...wrong answer to this equation. The answer is {a = -2, z = z, x = -3*z-2, y = -5*z-3}, {z = z, a = (z+2)/z, x = 0, y = 1+z}
 

mikhail

Senior Member
Jan 24, 2003
9,575
#4
++ [ originally posted by Loppan ] ++
I have these equations, I need to figure out how to show how I got the answer that a could be 2 if x=2, y=0 and b=b ( I ran it in a computer program) I know I have to use Gauss elinimination but how??? :

{x-y+a*z=1, 2*x-y+z=-1, a*x+y-z=1};

Form a matrix of the coefficients of x, y &z
(1 -1 a) (x) = (1)
(2 -1 1) (y) (-1)
(a 1 -1) (z) (1)

Take multiples of the top line from the later ones to get a zero entry in the left hand entry of the other two, and then take a multiple of the (new) middle line from the (new) last line to get a zero entry in the second entry there too.

Now the last line gives (Something messy)*z=(something else)
so you get z. Sub for z in the second line to solve for y, and then sub for y and z in the top line to get x.
 

gray

Senior Member
Moderator
Apr 22, 2003
30,098
#5
Set the matrices as shown above, and using matrix multiplications, rewrite the linear system above as the matrix equation

X . Z = Y.

This is nicer than the messy equations, right?. Now the matrix X is called the matrix coefficient of the linear system. The matrix Y is called the nonhomogeneous term. The matrix Z is the unknown matrix. Its entries are the unknowns of the linear system. The augmented matrix associated with the system is the matrix [X|Z], where

[X]Z] = (messy stuff here)

I hope I explained it okay...it's a bit complicated to explain online :down:
 

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