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Find the value(s) of a real parameter k so that the system of equations has infinitely many solutions.

$x+y+z=1$
$2x+ky+3z=-2$
$3x+5y+kz=-1$

Please help, I don't know how to go about this. Would you have to try every combination of the equations until you get a contradiction?

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2 Answers

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Add the first and second and subtract the third equation we have: $(k-4)(y-z) = 0$. Thus we require $k = 4$ to have infinite solutions.

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Firs you rewrite the system of equation as a matrix equation $Ax=b$. The system has a unique solution if and only if $\det(A)\neq0$. But since $\det A=(k-4)(k-1)=0$. So the solution is unique if and only if $k\neq 4$ and $k\neq 1$.

But if you check it further, when $k=1$ the system has no solution while when $k=4$ the system has infinitely many solutions. So $k=4$ is the only answer.

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