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

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(D) The given system of equations is:$kx + y + z = 1$$x + ky + z = k$$x + y + kz = k^2$First,we calculate the determinant $\Delta$ of the coefficient

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$-2$

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(D) The given system of equations is:$kx + y + z = 1$$x + ky + z = k$$x + y + kz = k^2$First,we calculate the determinant $\Delta$ of the coefficient matrix:$\Delta = \begin{vmatrix} k & 1 & 1 \ 1 & k & 1 \ 1 & 1 & k \end{vmatrix}$$= k(k^2 - 1) - 1(k - 1) + 1(1 - k)$$= k(k - 1)(k + 1) - (k - 1) - (k - 1)$$= (k - 1)[k(k + 1) - 1 - 1]$$= (k - 1)(k^2 + k - 2)$$= (k - 1)(k + 2)(k - 1) = (k - 1)^2(k + 2)$For the system to have no solution,we require $\Delta = 0$ and at least one of the Cramer's determinants $(\Delta_1, \Delta_2, \Delta_3)$ to be non-zero.If $k = 1$,the equations become $x + y + z = 1$,$x + y + z = 1$,and $x + y + z = 1$. This represents the same plane,so there are infinitely many solutions.If $k = -2$,$\Delta = 0$. Let us check $\Delta_1$:$\Delta_1 = \begin{vmatrix} 1 & 1 & 1 \ -2 & -2 & 1 \ 4 & 1 & -2 \end{vmatrix} = 1(4 - 1) - 1(4 - 4) + 1(-2 + 8) = 3 - 0 + 6 = 9 
eq 0$.Since $\Delta = 0$ and $\Delta_1 
eq 0$,the system has no solution when $k = -2$.

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

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