SOLUTION: Use matrices to solve the system. (If the system has infinitely many solutions, express your answer in terms of c, where x = x(c), y = y(c), and z = c. If the system has no solutio

Algebra ->  Matrices-and-determiminant -> SOLUTION: Use matrices to solve the system. (If the system has infinitely many solutions, express your answer in terms of c, where x = x(c), y = y(c), and z = c. If the system has no solutio      Log On


   

Question 661767: Use matrices to solve the system. (If the system has infinitely many solutions, express your answer in terms of c, where x = x(c), y = y(c), and z = c. If the system has no solution, enter NONE for each answer.)
2x-3y+z=-3
3x+2y-z=1
5x-2y+z=5
(x,y,z)=?
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--

Your three equations are: 
2x-3y+z=-3
3x+2y-z=1
5x-2y+z=5

The 3x4 augmented matrix is

2, -3, 1, -3
3, 2, -1, 1
5, -2, 1, 5

Now we perform row operations to attempt to transform the matrix into reduce row echelon 
form (rref).

Step one. We want the first element in Row 1 to be a 1. We multiply the entire row by 1/2 because 1/2 times 2 is 1.

Multiply (1/2)*R1.

1, -3/2, 1/2, -3/2
3, 2, -1, 1
5, -2, 1, 5

Now we want the first element in Row 2 to be 0. We add -3 times Row 1 to Row 2.

Add -3*R1 to R2.

1, -3/2, 1/2, -3/2
0, 13/2, -5/2
5,-2, 1, 5

Next, we want the first element in Row 3 to be 0. We add -5 times Row 1 to Row 3.

Add -5*R1 to R3.

1, -3/2, 1/2, -3/2
0, 13/2, -5/2, 11/2
0, 11/2, -3/2, 25/2

Notice that Column 1 looks like 1 0 0. We Want the 2nd Column to look like 0 1 0 so we repeat 
the process we just performed on the first column.

Multiply (2/13)*R2.

1, -3/2, 1/2, -3/2
0, 1, -5/13, 11/13
0, 11/2, -3/2, 25/2

Add (3/2)*R2 to R1.

1, 0, -1/13, -3/13
0, 1, -5/13, 11/13
0, 11/2, -3/2, 25/2

Add (-11/2)*R2 to R3.

1, 0, -1/13, -3/13
0, 1, -5/13, 11/13
0, 0, 8/13, 102/13

We want the third column to look like 0 0 1. We multiply every element in Row 3 by 13/8 
because (8/13)*(13/8)=1. Then repeat the above process to get zeroes in the other column 3 
positions.

Multiply (13/8)*R3

1, 0 -1/13, -3/13
0, 1, -5/13, 11/13
0, 0, 1, 51/4

Add (5/13)*R3 to R2.

1, 0, -1/13, -3/13
0, 1, 0, 23/4
0, 0, 1, 51/4

Add (1/13)*R3 to R1.

1, 0, 0, 3/4
0, 1, 0, 23/4
0, 0, 1, 51/4

The matrix is now in rref. (x, y, z) = (3/4, 23/4, 51/4)

Were is a link to a row operation calculator website where you can enter your matrix and 
experiment with row operations to find the solution. This is excellent practice. You might 
considering trying it with the two matrix problems you submitted here so you can see how 
nicely it works.

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=roc

Best,
Mrs. Figgy