SOLUTION: Could someone assist in explaining how to correctly solve the question. My previous answer was -40-x-3z/2. Solve the system by Gaussian elimination. (If there is no solution, en

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Question 1129997: Could someone assist in explaining how to correctly solve the question. My previous answer was -40-x-3z/2.
Solve the system by Gaussian elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, enter a general solution in terms of one of the variables.)
0.1x − 0.2y + 0.3z = 4
0.5x − 0.1y + 0.4z = 16
0.7x − 0.2y + 0.3z = 16

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Your "answer" makes no sense; the answer should show the values of x, y, and z that satisfy all three equations.

There are always an endless number of different ways to solve a system of equations using Gaussian elimination. I will show one.

The beginning matrix is this:

matrix%283%2C4%2C0.1%2C-0.2%2C0.3%2C4%2C0.5%2C-0.1%2C0.4%2C16%2C0.7%2C-0.2%2C0.3%2C16%29

The first thing I would do is multiply each row (i.e., each equation) by 10 to get rid of the decimals, since working with integers is much easier.

matrix%283%2C4%2C1%2C-2%2C3%2C40%2C5%2C-1%2C4%2C160%2C7%2C-2%2C3%2C160%29

Now before I plunge into the standard techniques for Gaussian elimination, I'm going to look at the equations to see if there is some major simplification that can be made. And I see that the coefficients of y and z in both the first and third equations are the same (-2 and 3); that means I can get an equation in x only by subtracting the first equation from the third.

So my first step (after converting the matrix to all integers) will be to replace row 3 with (row 3 minus row 1):

matrix%283%2C4%2C1%2C-2%2C3%2C40%2C5%2C-1%2C4%2C160%2C6%2C0%2C0%2C120%29

Next I will combine two steps: dividing row 3 by 6, and making it the first row (since it will be in the form I want for the first row of the final matrix):

matrix%283%2C4%2C1%2C0%2C0%2C20%2C1%2C-2%2C3%2C40%2C5%2C-1%2C4%2C160%29

Next use row 1 to get 0's in the first column of rows 2 and 3: replace row 2 with (row 1 minus row 2) and replace row 3 with (row 3 minus 5 times row 1):

matrix%283%2C4%2C1%2C0%2C0%2C20%2C0%2C2%2C-3%2C-20%2C0%2C-1%2C4%2C60%29

The first column of my matrix is in its final form; next I want to get a "1" in row 2 column 2. I can do that by replacing row 2 with (row 2 plus row 3):

matrix%283%2C4%2C1%2C0%2C0%2C20%2C0%2C1%2C1%2C40%2C0%2C-1%2C4%2C60%29

Next, to finish column 2, I need a "0" in row 3 column 2; I can do that by replacing row 3 with (row 3 plus row 2):

matrix%283%2C4%2C1%2C0%2C0%2C20%2C0%2C1%2C1%2C40%2C0%2C0%2C5%2C100%29

Column 2 is finished; next I need a "1" in row 3 column 3. I can get that by dividing row 3 by 5:

matrix%283%2C4%2C1%2C0%2C0%2C20%2C0%2C1%2C1%2C40%2C0%2C0%2C1%2C20%29

And last I need a "0" in row 2 column 3; I get that by replacing row 2 with (row 2 minus row 3):

matrix%283%2C4%2C1%2C0%2C0%2C20%2C0%2C1%2C0%2C20%2C0%2C0%2C1%2C20%29

The matrix is in the desired final form; it shows the solution to the system is

x=20; y=20; z=20