SOLUTION: I need to use the Gaussian elimination method on the following: For a circuit, the currents i1, i2, and i3 measured in amperes, are determined by the following linear equations

Question 967753: I need to use the Gaussian elimination method on the following:

For a circuit, the currents i1, i2, and i3 measured in amperes, are determined by the following linear equations:
-i1 + 4i3 = 3
5i1 - 3i2 = 26
-3i1 +6i2 = -21
Any help would be very helpful.
Thanks in advance

Answer by ikleyn(52879)   (Show Source): You can put this solution on YOUR website!

Let's introduce new variables x = i1, y=i2 and z=i3 for brevity.
Then you need to solve the system of three linear equations in three unknowns

-x   +         4z = 3,
5x   - 3y         = 26,
-3x + 6y         = -21

by the Gauss's elimination method.

Notice that your system just contains the sub-system of two linear equations in two unknowns x and y. It is the system consisting of the second and the third equations.
Its facilitates the solution.

So, we need to solve this sub-system, which is

5x   - 3y = 26,
-3x + 6y = -21.

Multiply the first equation of these two by 2 and add to the second equation to eliminate y. You will get

    10x   - 6y = 52,
+
     -3x + 6y = -21.
--------------------
      7x          = 31.

It gives you the solution for x: x = .

Next, substitute it to any of the second or the third equation of the original system. It gives you the solution for y.

Then substitute the found values of x and y into the first equation, and you will get the solution for z.

Good luck.

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Comment from student: Hi ikleyn, thanks for the response. How would I solve this question using an augmented matrix and using gaus elimination? Thanks again for the help.
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OK, let's do this.

Your augmented matrix is
.

Multiply the first row by 5 and add to the second row to eliminate x in the second equation. You will get
.

Next, multiply the first row by -3 and add to the third row to eliminate x in the third equation. You will get
.

OK, we just made the zeroes in the first column below "-1".

Now, multiply the second row of the last matrix by 2 and add to the third row to eliminate y in the third equation. You will get
.

Thus you just completed the elimination process: you transformed your original matrix to the upper triangle matrix.
Hence, the solution for z is z = = = .

Now you should make back-substitution.
Substitute the found value of = into the equation
-3y + 20z = 145   (in accordance with the second row of the last matrix). It will give you the solution for .

Then back-substitute the found values of    and z into the very first equation and find x.

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