how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0
I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it.
You create a matrix using the constants in the equations. Below, you will see
approximately what the matrix would look like. Unfortunately, it is impossible
to get it to display properly on this page.
┌ ┐
│ 4 0 1 | 3 │
│ 2 -1 0 | 2 │
│ 0 3 2 | 0 │
└ ┘
The idea is to perform the same math procedures on these
matrix rows that you would perform on the full equations. We
want the first row to be 1 0 0 | a, meaning that whatever value
appears as the a entry is the value of x. The second row
has to be 0 1 0 | b, and the third row has to be 0 0 1 | c.
Multiply row 2 by 2.
┌ ┐
│ 4 0 1 | 3 │
│ 4 -2 0 | 4 │
│ 0 3 2 | 0 │
└ ┘
Now subtract row 2 from row 1, and replace row 2 with the result.
┌ ┐
│ 4 0 1 | 3 │
│ 0 2 1 | -1 │
│ 0 3 2 | 0 │
└ ┘
Multiply row 2 by 2.
┌ ┐
│ 4 0 1 | 3 │
│ 0 4 2 | -2 │
│ 0 3 2 | 0 │
└ ┘
Subtract row 3 from row 2, replacing row 2.
┌ ┐
│ 4 0 1 | 3 │
│ 0 1 0 | -2 │
│ 0 3 2 | 0 │
└ ┘
Multiply row 2 by 3 and subtract row 3, replacing row 3.
┌ ┐
│ 4 0 1 | 3 │
│ 0 1 0 | -2 │
│ 0 0 -2 | -6 │
└ ┘
Divide row 3 by -2.
┌ ┐
│ 4 0 1 | 3 │
│ 0 1 0 | -2 │
│ 0 0 1 | 3 │
└ ┘
Subtract row 3 from row 1, replacing row 1.
┌ ┐
│ 4 0 0 | 0 │
│ 0 1 0 | -2 │
│ 0 0 1 | 3 │
└ ┘
Divide row 1 by 4.
┌ ┐
│ 1 0 0 | 0 │
│ 0 1 0 | -2 │
│ 0 0 1 | 3 │
└ ┘
Row 1 shows that x = 0
Row 2 shows that y = -2
Row 3 shows that z = 3
Plug those values into the original equations to check the answer.
4x + z = 3
4(0) + 3 = 3
0 + 3 = 3
3 = 3
2x – y = 2
2(0) – (-2) = 2
0 + 2 = 2
2 = 2
3y + 2z = 0
3(-2) + 2(3) = 0
-6 + 6 = 0
0 = 0
