Solve this system equation by row reduction.?

Question

x+y-z=4
2x-y=3
5x-y-z=6

Answer 1

x + y - z = 4
2x - y = 3
5x - y - z = 6

Your first step would be to convert this into a matrix.

[1 1 -1 | 4]
[2 -1 0 | 3]
[5 -1 -1 | 6]

To row reduce this, perform the following operations:
R2 -> R2 - 2R1 {subtract 2 times the first row from the 2nd row}
R3 -> R3 - 5R1 {subtract 5 times the first row from the 3rd row}.

Remember your goal is to put this into reduced row echelon form, and that means putting 0s under the leading 1.

[1 1 -1 | 4]
[0 -3 2 | -5]
[0 -6 4 | -14]

Now, you require a leading 1 in the second row.

R2 -> (-1/3) R2

[1 1 -1 | 4]
[0 1 -2/3 | 5/3]
[0 -6 4 | -14]

Now you want *this* leading 1 surrounded by 0s. This is accomplished by:

R1 -> R1 - R2
R3 -> R3 + 6R2

[1 0 1/3 | 1]
[0 1 -2/3 | 5/3]
[0 0 0 | -4]

At this point, since we cannot form the identity matrix, it follows we have no solution.

Answer 2

Look at this like a matrix:

x + y - z = 4 is : [1 1 -1 4]
2x - y = 3 is : [2 -1 0 3]
5x - y -z = 6 is : [5 -1 -1 6]

When you row reduce this matrix, you'll notice that it is singular...
This means you can't row reduce it all the way.

Answer 3

row reduction?

just use a systems of equations.
substitution is best!

Answer 4

first delete commonalities with the 3 variable ones.

Answer 5

**** thats hard

no help srry

Answer 6

umm i think whether u r stupid or lazy i'm sorry but it's true k? ask ur teacher .....

Answer 7

those are to easy to answer