The system of equationkx+y+z=1,x+ky+z=k  and x+y+kz=k^{^{2}} have resolution if k equals

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Answers (1)
V Vakul

As learnt in

Solution of a system of equations -

x_{1},x_{2},\cdot \cdot \cdot ,x_{n} satisfy the system of linear equations  A_{x}=B

- wherein

 

 

Determinant is \begin{vmatrix} k &1 & 1\\ 1& k &1 \\ 1& 1 &k \end{vmatrix}

 

on solving, k(k^{2}-1)-1(k-1)+1(1-k)

=k^{3}-k-k+1+1-k

=k^{3}-3k+2

For     k=1

k^3-k^2+k^2-k-2k+2

\Rightarrow k^{2}(k-1)+k(k-1)-2(k-1)

\Rightarrow (k-1)[k^{2}+2k-k-2]

\Rightarrow (k-1)(k-1)(k+2)

It has a solution when k+2=0

k=-2

and \Delta_{1}, \Delta_{2}, \Delta_{3} are all zero for k=-2

 

 


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