Question

Help with Algebra 2?

Answer 1

Solve system by elimination.

Answer 2

@IAmSinged

Answer 3

x = -3, y = 6, z = -1 using a matrix solver I wrote.
Check it using : http://www.bluebit.gr/matrix-calculator/linear_equations.aspx

Answer 4

*x = 3, typo'd

Answer 5

Im sorry I dont get it

Answer 6

can you explain it a little more please?

Answer 7

@zepdrix

Answer 8

@IAmSinged

Answer 9

I solved the matrix using very high-level math, so I'll tell you a "dumber" way to solve it.

The method is called Naive Gaussian Elimination, because if there are any numbers on the diagonal, it will fail completely.

First, we turn the system into a matrix system Ax = b:
A:
-1 -1 -1
-4 4 5
2 0 2

B:
-8
7
4

A is me taking everything before the equal sign and putting it into a matrix, and B is me taking everything right of the equal sign and putting it into a matrix.
posting this now for you to have something to look at

Answer 10

Continuing:

So we take -1 from the first row of the matrix (aka our first equation) and what we want to do is to zero out the -4 and 2 below it.
*NOTE: row and equation mean the same thing.

How do we do this? We do this by adding row 1 multiplied by some constant to the second and third rows. What is this constant? This constant is whatever number multiplied by -1 will get me -4 in the case of 0'ing the 2nd row and whatever number multiplied by -1 gets us 2.

For row 2 in our row matrix:
We know that 4 * -1 = -4, so we know that our constant is 4. This tells us we are going to subtract row 1 from row 2 4 times.
So we replace row 2 with:
[-4 4 5] - 4[ -1 -1 -1] = [0 8 9]
We do the same for our matrix B:
7 - (4 * -8) = 39

So now our A matrix is:
-1 -1 -1
0 8 9
2 0 2

and our B matrix is:
-8
39
4

Answer 11

Continuing:
For row 3:
We know that -2 * -1 = 2, so we are going to subtract -1 * row 1 from row 3 twice, which basically means we're adding 2* row 1 to row 3.

So for matrix A we do:
row 3 = [ 2 0 2] + 2[ -1 -1 -1] = [0 -2 0]

likewise, we do the same for matrix B:
row 3 = 4 - 16 = -12

So now our matrix A is:
-1 -1 -1
0 8 9
0 -2 0

And our B matrix is:
-8
39
-2

This is equivalent to saying that our system of equations is now:
-x-y-z = -8
8y + 9z = 39
-2y = -12

We ignore the 3rd element in the diagonal, because under the last element of a diagonal in a square matrix, there is nothing below it to 0 out.
Note for that specific case we have an equation with 1 variable, which is -2y = -12.
This gets us y = 6.
Substituting y =6 into the second equation which has 2 variables, we get:
48 + 9z = 39
9z = -9
z = -1

So now we know that y = 6 and z = -1. Plugging into the first equation, we get x = 3.
So now we have all 3 variables:
x = 3, y =6, z = -1

This system of equations is kind of weird for NGE, since we have 0 on the diagonal, which normally means NGE fails. However, we ended up with an equation that only has y, so it ended up working out.

A video is probably much better than a wall of text if you mind just shuts down from looking at walls of text:
NGE in general:
https://www.youtube.com/watch?v=m3ooPCNaoTA