The system of equations kx + y + z =1 x + ky | vedclass

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Question

$k x+y+z=1$

$x+k y+z=k$

$x+y+z k=k^{2}$

$\Delta=\left|\begin{array}{ccc} K & 1 & 1 \ 1 & K & 1 \ 1 & 1 & K \end{arr

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

$k x+y+z=1$

$x+k y+z=k$

$x+y+z k=k^{2}$

$\Delta=\left|\begin{array}{ccc} K & 1 & 1 \ 1 & K & 1 \ 1 & 1 & K \end{array}ight|= K \left( K ^{2}-1ight)-1( K -1)+1(1- K )$

$= K ^{3}- K - K +1+1- K$

$= K ^{3}-3 K +2$

$=( K -1)^{2}( K +2)$

For $K =1$

$\Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$

But for $K =-2,$ at least one out of $\Delta_{1}, \Delta_{2}, \Delta_{3}$ are not zero

Hence for no sol$^{n}$, $K =-2$

The system of equations $kx + y + z =1$ $x + ky + z = k$ and $x + y + zk = k ^{2}$ has no solution if $k$ is equal to

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