Math question?

Question

solve the system of 3 equations:

x+2y+3z=0
2x+3y+2z=3
-x+y+2z=-2

I asked this one earlier but wasn't exactly looking for the answer, I really need to know how to do this... I get lost with the books explanation (Gauss Method) and need some real help.

Answer 1

The corresponding matrix is

1 2 3 0
2 3 2 3
-1 1 2 -2

Want to have zeros underneath the main diagonal.
First we "manufacture" zeros in the first column, in the rows 2 and 3.

R2 -> R2 - 2R1 (subtract twice row 1 from row 2)
R3 -> R3 + R1 (add row 1 to row 3)

1 2 3 0
0-1-4 3
0 3 5 -2

multiply R2 by (-1)

1 2 3 0
0 1 4 -3
0 3 5 -2

Now we "manufacture" zeros in the second column, row 3.

R3 -> R3 - 3R2

1 2 3 0
0 1 4 -3
0 0 -7 7

Divide row 3 by (-7)

1 2 3 0
0 1 4 -3
0 0 1 -1

You can now continue to obtain ceros also ABOVE the main diagonal, but this is not crucial.

In the bottom row we have
z = -1

In the second row :
y+4z = -3 => y -4 = -3 => y = 1

In the top row:
x + 2y + 3z = 0
x + 2 - 3 = 0 => x = 1

The solution of the system:
(x, y, z) = (1, 1, -1)

Hope this helps.

Answer 2

Okay, let's first solve two of the equations for the same variable. For sake of simplicity, let's choose to solve the first and third equations for x.

From the first equation: x = -2y -3z
From the third equation: x = y + 2z + 2

Hence, it follows that:
-2y -3z = y + 2z + 2
Simplifying this, we get:
3y + 5z = -2
... and we have eliminated x, giving us one equation with the two unknowns y and z.

Now let's eliminate x from a different pair of equations.
Let's go with the first and second equations.

From the first equation: x = -2y -3z
For sake of easy numbers, let's double this:
2x = -4y -6z
The second equation is 2x + 3y +2z = 3
Plugging in our earlier solution for 2x, we get:
(-4y -6z) + 3y + 2z = 3
which simplifies to
y + 4z = -3

So now you have two equations with two unknowns:
3y + 5z = -2
y + 4z = -3

... and I presume you know how to solve two equations with two unknowns, so you can take the problem from here.

The real trick to the problem is to eliminate variables: Take one pair of equations and put them together to eliminate a variable. Then when you've done that, take a different pair of equations, and eliminate the *same* variable as you did for the first pair.

Hopefully that helps!

Answer 3

There are many ways to solve systems of equations. Later in calculus I used matrices for this type of problem to do it quickly. However, the basic method is (i believe) termed the Gaussian Method. It works because we have three equations and three unknowns. Solve one equation for x, solve another for y, and plug these into the third equation to get z. Repeat the process to get x and y.