SOLUTION: For what value of k does the following system have infinitely many solutions? kx + y + z=0 x + 2y + kz = 0 -X+3z=0

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Question 1198864: For what value of k does the following system have infinitely many solutions?
kx + y + z=0
x + 2y + kz = 0
-X+3z=0
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Solve the 3rd equation for z and substitute in the other two equations.

x = 3z

3kz+y+z=0
3z+2y+kz=0

y+(3k+1)z=0
2y+(3+k)z=0

There are infinitely many solutions if the two equations are equivalent.

2y+(6k+2)z=0
2y+(3+k)z=0

6k+2=3+k
5k=1
k=1/5

ANSWER: k=1/5

Answer by ikleyn(52785) About Me  (Show Source):
You can put this solution on YOUR website!
.

This system has infinitely many solutions if and only if the determinant of the
coefficient matrix is zero.

So, calculate the determinant of the matrix as a function of "k" and equate it to zero.

From this equation, find the value of "k".