Solve the given linear system

[frctl]

2x - y - 3z = 0
-x + 2y - 3z = 0
x + y + 4z = 0
augmented matrix
2 1 -3 0
-1 2 -3 0
1 1 4 0

How can I eliminate the 2 and the -1?

 

[Deleted member 4993]

2x - y - 3z = 0
-x + 2y - 3z = 0
x + y + 4z = 0
augmented matrix
2 1 -3 0
-1 2 -3 0
1 1 4 0

How can I eliminate the 2 and the -1?

Why do you want to do that? From where do you want to eliminate '2'?

 

[frctl]

I normally solve the system by reducing to rref

 

[frctl]

I can swap row1 with row3 since I want a zero in the first column.

 

[Deleted member 4993]

I normally solve the system by reducing to rref

First divide R1 by 2 - so that [1,1] becomes 1 [pivot].

then add resulting (new) R1 to R2 so that [2,1] becomes 0.

 

[Deleted member 4993]

I can swap row1 with row3 since I want a zero in the first column.

Yes you can do that too. Do the operation with which you are comfortable!

 

[frctl]

Then I obtain fractions:
1 1/2 -3/2 0

 

[frctl]

augmented matrix
2 1 -3 0
-1 2 -3 0
1 1 4 0
swap R1 and R3
1 1 4 0
-1 2 -3 0
2 1 -3 0

I wish to get rid of -1 in R2

 

[frctl]

correction
augmented matrix
2 -1 -3 0
-1 2 -3 0
1 1 4 0
swap R1 and R3
1 1 4 0
-1 2 -3 0
2 1 -3 0

 

[Deleted member 4993]

correction
augmented matrix
2 -1 -3 0
-1 2 -3 0
1 1 4 0
swap R1 and R3
1 1 4 0
-1 2 -3 0
2 1 -3 0

Good ... where are you stuck?

 

[frctl]

Knowing which further operations are required.

 

[Deleted member 4993]

correction
augmented matrix
2 -1 -3 0
-1 2 -3 0
1 1 4 0
swap R1 and R3
1 1 4 0
-1 2 -3 0
2 1 -3 0

You have been doing similar problems with matrices! What did you learn from those!?

1 1 4 0
-1 2 -3 0
2 1 -3 0

You want to add R1 into R2 to get:

1 1 4 0
0 3 1 0
2 1 -3 0

Then multiply R1 by 2 and subtract from R3 to get:

1 ... 1 .... 4 ... 0
0 ... 3 .... 1 ... 0
0 .. -1 . -11 ... 0

Why did I take these steps?

What is the next step?

 

[frctl]

To eliminate the 3 in row2

 

[Deleted member 4993]

To eliminate the 3 in row2

Are you talking about next step?

You did not tell us about your "reasons" for the steps that was shown before!

Anyway, the next step could be to divide R2 by 3, so that your pivot number [2,2] becomes 1

1 ... 1 .... 4 ... 0
0 ... 1 ..(1/3) . 0
0 .. -1 . -11 ... 0

What would you do next? If you do not know - review the problems you have solved before.

 

[frctl]

by multiplying row2 by 1/3 ?

 

[Deleted member 4993]

by multiplying row2 by 1/3 ?

Why would you want to do that?

Do it and see where it takes you!

Are you taking a regular math-class or doing these on your own?

 

[frctl]

on my own

 

[Steven G]

Regardless of what textbooks and instructors say you do NOT have to get 1's in the leading positions untill the very last step. This way you do not have to deal with fractions until the last step. If you have done your rrer with the exception of getting 1's on the diagonals, then the last step would be to divide by the non-zero leading numbers. Try it you'll like it

 

[Otis]

Are you finished with this exercise, frctl?