math problem?

Question

write the augmented matrix of each system of equations.

{2x - y - z = 0
{x - y + z = 1
{ 3x - y = 2

can you explain in detail how you solve this?

Answer 1

Hi,

The augmented matrix is:

[2..-1..-1...0]
[1..-1....1..1]
[3..-1....0..2]

The easiest way to solve this on a TI83 calculator is to press MATRIX, MATH, and select B: rref( which will give you reduced row echelon form. After rref( appears on your screen, press matrix and select the name of the matrix where you entered this 3 x 4 matrix. If it was [A], your screen says rref([A], you will see:

[1..0..0..1]
[0..1..0..1]
[0..0..1..1]

This means x = 1, y = 1, and z = 1.

I hope this helps!! :-)

Answer 2

2 -1 -1 0 <-- Row 1
1 -1 1 1 <-- Row 2
3 1 0 2 <-- Row 3

You object is to manipulate the above matrix so that it looks like The below matrix where x1, y1, and z1 are the answers.
1 0 0 x1
0 1 0 y1
0 0 1 z1
For example, if you replace Row1 with Row2 - Row1 the matrix original becomes:
1 0 2 -1
1 -1 1 1
3 -1 0 2
Just continue manipulations until you get the second matrix shown above. The 3X3 matix will have the diagonal = all ones, and zroes everywhere else. Start with the ist column and continuue manipulations until the column is 1,0,0.
Then go to column 2, and then to column 3.

Answer 3

from third equation . . . y = 3x -2 . substitute to first two equation
2x - 3x +2 - z = 0 . . from equation 1
-x -z + 2 = 0
x +z - 2 = 0 . . . . equation 1

x - 3x + 2 +z = 1
- 2x + z + 1 = 0 . . . . . equation 2
x + z - 2 = 0 . . . . equation 1
subtract the 2 equation
-3x + 3 = 0
x = 1
y = 3x -2 = 1
z = 1

Answer 4

2x - y - z = 0
x - y + z = 1
-----------------
3x - 2y = 1

3x = 2y + 1

(2y +1) - y = 1
y + 1 = 1
y = 0

3x - 0 = 2
3x = 2
x = 2/3

2(2/3) - 0 - z = 0
4/3 - z = 0
z = 4/3