# Thread: Two lin alg question

1. ## Two lin alg question

1. Use gauss-jordan elimination to solve the following systems
x - 2y + z = -6
2x - 3y = 7
-x + 3y - 3z =11
I got a finite solution for this one which was wrong so I'm redoing it. Is it suppose to be infinite?

Ok. so I got x-3z=4
y-2z=5
and 0=0
Now im stuck.

2. Determine the polynomial function whose graph passes through the points: (1,4), (2,0), (3,12)
This one i'm stuck on. Curve fitting?

2. Originally Posted by Db0ne
1. Use gauss-jordan elimination to solve the following systems
x - 2y + z = -6
2x - 3y = 7
-x + 3y - 3z =11
I got a finite solution for this one which was wrong so I'm redoing it. Is it suppose to be infinite?

Ok. so I got x-3z=4
y-2z=5
and 0=0
Now im stuck.

2. Determine the polynomial function whose graph passes through the points: (1,4), (2,0), (3,12)
This one i'm stuck on. Curve fitting?
1. Since the determinant of the coefficient matrix is equal to zero, there is either no solution or an infinite number of solutions. I haven't checked your work but assuming that x - 3z = 4 and y - 2z = 5 are correct, the solution can be written as

x = 3t + 4, y = 2t + 5, z = t

where t is any real number. That is, an infinite number of solutions.

2. Assuming that a polynomial of minimum degree is required, you should use the model $\displaystyle y = ax^2 + bx + c$ and substitute the given data to get three equations involving a, b and c. Solve these equations simultaneously.

3. Originally Posted by Db0ne
1. Use gauss-jordan elimination to solve the following systems
x - 2y + z = -6
2x - 3y = 7
-x + 3y - 3z =11
I got a finite solution for this one which was wrong so I'm redoing it. Is it suppose to be infinite?

Ok. so I got x-3z=4
Then you've added wrong! Or, perhaps, subtracted where you should have added. Adding the last two equations cancels the "y" term : 2x-x- 3z= x- 3z= 7+ 11= 18, not 4.

y-2z=5
and 0=0
Now im stuck.
Since x- 3z= 18, x= 3z+ 18 and the first equation becomes 3z+ 18- 2y+ z= -6 so 4z- 2y= -24. Dividing by 2, 2z- y= -12 or y= 2z+ 12.

Putting x= 3z+ 18 and y= 2z+ 12 into the last equation, -(3z+ 18)+ 3(2z+ 12)- 3z= 11 or -3z+ 6z- 3z- 18+ 36= 0z+ 18= 11 which is never true. The is no x, y, z that satisfy these three equations.

2. Determine the polynomial function whose graph passes through the points: (1,4), (2,0), (3,12)
This one i'm stuck on. Curve fitting?
A quadratic function will fit any three points (in general, a function of degree n-1 will fit n points). Let y= $\displaystyle ax^2+ bx+ c$ and put in those three points:
at (1, 4): $\displaystyle 4= a(1^2)+ b(1)+ c$ so a+ b+ c= 4
at (2, 0):$\displaystyle 0= a(2^2)+ b(2)+ c$ so 4a+ 2b+ c= 0
at (3 12): $\displaystyle 12= a(3^2)+ b(3)+ c$ so 9a+ 3b+ c= 12.

Solve those three equations for a, b, and c.