Hey guys I just want to double check if I have the right answer to a problem. Thanks in advance :)

Use row-echelon form of matrices to solve the system.

2x+4z=1

x+y+3z=0

x+3y+5z=0

I got the answer (1/2-a,-a,a) where a is any real number

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- Oct 27th 2009, 10:39 AMcornearsRow Echelon
Hey guys I just want to double check if I have the right answer to a problem. Thanks in advance :)

Use row-echelon form of matrices to solve the system.

2x+4z=1

x+y+3z=0

x+3y+5z=0

I got the answer (1/2-a,-a,a) where a is any real number - Oct 28th 2009, 08:56 AMHallsofIvy
Why do you need us to check it? It is just arithmetic.

If x= 1/2- a, y= -a, and z= a then the first equation becomes 2(1/2- a)+ 4(a)= 1- 2a+ a= 1- a. No, that is not correct so this is not a solution.

Or did you mean x= 1/(2-a), y=-a, z= a? In that case, 2x+ 4z= 2/(2- a)+ 4a which is still not 1.

No, sorry, your solution is not correct.