SOLUTION: I am not understanding how to Solve the system of linear equation using matrices 4x-y+2z=5 2y+z=4 4x+y+3z=10 Can you guide me step by step in solving it?

4x-y+2z=5
  2y+z=4
4x+y+3z=10

 4x-1y+2z =  5
 0x+2y+1z =  4
 4x+1y+3z = 10

Erase all the letters

 4 -1 +2 =  5
 0 +2 +1 =  4
 4 +1 +3 = 10

Erase all the + signs

 4 -1  2 =  5
 0  2  1 =  4
 4  1  3 = 10

Replace the equal signs with a |

⎾4 -1  2 |  5⏋
⎢0  2  1 |  4⎢
⎿4  1  3 | 10⏌

We want to end up with a matrix that looks like this:

⎾1  0  0 | ⎽⏋
⎢0  1  0 | ⎽⎢
⎿0  0  1 | ⎽⏌

-------------------------

First we get all the 0's

⎾4 -1  2 |  5⏋
⎢0  2  1 |  4⎢
⎿4  1  3 | 10⏌

To get a 0 where the 4 is on the lower left hand corner,
multiply the top row by -1

⎾-4  1 -2 | -5⏋
⎢ 0  2  1 |  4⎢
⎿ 4  1  3 | 10⏌

Add the first row to the bottom row:

⎾-4  1 -2 | -5⏋
⎢ 0  2  1 |  4⎢
⎿ 0  2  1 |  5⏌

To get a 0 where the 1 is in the top row, multiply
the top row by -2

⎾ 8 -2  4 | 10⏋
⎢ 0  2  1 |  4⎢
⎿ 0  2  1 |  5⏌
 
Add the middle row to the top row:

⎾ 8  0  5 | 14⏋
⎢ 0  2  1 |  4⎢
⎿ 0  2  1 |  5⏌

To get a 0 where the 2 on the bottom row is, multiply
the bottom row by -1

⎾ 8  0  5 | 14⏋
⎢ 0  2  1 |  4⎢
⎿ 0 -2 -1 | -5⏌

Add the middle row to the bottom row

⎾ 8  0  5 | 14⏋
⎢ 0  2  1 |  4⎢
⎿ 0  0  0 | -1⏌

If you replace the letters now 

  8x + 0y + 6z = 14
  0x + 2y + 1z =  4
  0x + 0y + 0z =  1

The bottom equation has no solution because the left
side is 0 and the right side is 1, so there can be no
solution.

Edwin

Hi

Note:  finding the value of a 3x3 matrix is done as follows:

 = a1(b2c3-c2b3) - b1(a2c3-c2a3) + c1(a2b3-b2a3) 

4x-y+2z=5

2y+z=4

4x+y+3z=10 

 



D =  = 0

    4*5 -(-1)1*-4 +2*-8 = 0

  this system has NO solution as D = 0