I have the following system: 2x - y + 7z = 0 -x + 2y -3z = 0 x + y + 4z = 0 I have determined the solution to be: x = -y -4t y = -1/3t z = t what I need to know now, is if (0, 7, -1) is a solution to the system?
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Originally Posted by blackhug I have the following system: 2x - y + 7z = 0 -x + 2y -3z = 0 x + y + 4z = 0 I have determined the solution to be: x = -y -4t y = -1/3t z = t what I need to know now, is if (0, 7, -1) is a solution to the system? x = 0 , y = 7 , and z = -1 makes the first equation of the system false ... so I'd say no.
lol yeah, I figured as much. I was just confused about it having infinite solutions. Thought it might have been a little more complicated than just putting it into the equation.
Originally Posted by blackhug I have the following system: 2x - y + 7z = 0 -x + 2y -3z = 0 x + y + 4z = 0 I have determined the solution to be: x = -y -4t y = -1/3t z = t what I need to know now, is if (0, 7, -1) is a solution to the system? Why is the red y part of your solution for x?
Originally Posted by mr fantastic Why is the red y part of your solution for x? Well because I was told to solve this system by reducing it to echelon form. So I was left with: 1 1 4 0 1 1/3 0 0 0 If i had my solution for x as wouldn't that be solving by reduced echelon form?
Originally Posted by mr fantastic Why is the red y part of your solution for x? Your solution for x, y and z needs to be in terms of the parameter t only.
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