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Solving simultaneous linear equation with three unknows is a damn knotty problem to me. Though, with the help of determinant, I can solve it effectively. But now, I don't wanna use determinants anymore. Thus, I've tried an alternative however to no avail. Pls. Prof. Treat me like a novice by solving the question below without using inverse matrix or determinants:

:( 3x+2y-z=19.........(1)

4x-y+2z=4...........(2)

2x+4y-5z=32....(3)

:(5x-3y-2z=31..........(1)

2x+6y+3z=4..........(2)

4x+2y-z=30............(3)

explain thoroughly how you solve it!

To solve each of them, I write down the problem in matrix format.

Once this has been done, I get a 1 in the 1st element of the 1st by dividing that row by 3.

I then get a zero in each cell below that by subtracting off the appropriate multiple of row 1.

For the 1st step, I divide row 1 by 3, subtract 4/3 for the 2nd row, and subtract 2/3 from the 3rd row. I also do it to the solutions. This gives

1 0.6667 -0.3333 6.3333

0 -3.6667 3.3333 -21.3333

0 2.6667 -4.3333 19.3333

In the next step, I multiply the 2nd row by -3/11. This puts a 1 in the place of -3.6667.

I multiply the 2nd row by 1/11 and add that to the 1st row. I multiply the 2nd row by 8/11 and add it the 3rd row. This gives a zero in the second column. The result is

1 0 0.2727 2.4545

0 1 -0.9091 5.8182

0 0 -1.9091 3.8182

To get a 1 in the 3rd position of the 3rd row, multiply that row by -11/21.

Add -10/21 of the 3rd row to the 2nd row. Add -1/7 of the 3rd row to the 1st row.

The solution turns out to be 3, 4, -2.

This is how to do the 2nd problem as well.

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Comment | Damn! To hell with your explanation! broadly speaking, I grasp nothing from ya Damn explanation. Expert or whatever you called yourself, go to yell!!! |

Algebra

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