x(k+2)+y(k-1)+z(k)=2 y(k+2)+2x=0 z(k^2+k-2)=k+2 Is there any val

Emmanuella Browne

Emmanuella Browne

Answered question

2022-02-18

x(k+2)+y(k1)+z(k)=2
y(k+2)+2x=0
z(k2+k2)=k+2
Is there any value of k for which this system of linear equations would have infinite solutions? I mean, it seems as if it does when k = -2, but it also seems as if it has no solutions when k = -2. Im

Answer & Explanation

Elijah Hunt

Elijah Hunt

Beginner2022-02-19Added 5 answers

If k=-2, the system has rank 2:
{3y2z=22z=0
hence the solutions are {(x,23,0)xR}. It does have an infinite set of solutions, the straight line (0,23,0)+x(1,0,0).
If k=1, the system also has rank 2:
{3x+z=23y+2z=00=3
but the last line is not identically 0, hence there is no solution.
On all other cases, the ystem has rank 3 and has a unique solution.

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