If the system of equations kx + 2y - z = 2, ( | vedclass

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(D) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D 
eq 0)$.The coeffic

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infinite

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(D) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D 
eq 0)$.The coefficient matrix is:$D = \begin{vmatrix} k & 2 & -1 \ k-1 & k & 1 \ 1 & k-1 & k \end{vmatrix}$Expanding the determinant along the first row:$D = k(k^2 - (k-1)) - 2(k(k-1) - 1) - 1((k-1)^2 - k)$$D = k(k^2 - k + 1) - 2(k^2 - k - 1) - (k^2 - 2k + 1 - k)$$D = k^3 - k^2 + k - 2k^2 + 2k + 2 - k^2 + 3k - 1$$D = k^3 - 4k^2 + 6k + 1$For a unique solution,$k^3 - 4k^2 + 6k + 1 
eq 0$.Let $f(k) = k^3 - 4k^2 + 6k + 1$. Since $f'(k) = 3k^2 - 8k + 6$,the discriminant of $f'(k)$ is $(-8)^2 - 4(3)(6) = 64 - 72 = -8 Since $f(k)$ is a cubic polynomial with real coefficients,it must have at least one real root. Let this root be $k_0$. The system has a unique solution for all $k \in \mathbb{R} \setminus \{k_0\}$.However,the question asks for the number of values of $k$ for which the system has a unique solution. Since there is only one value of $k$ (the root of $f(k)=0$) for which the system does not have a unique solution,there are infinitely many real values of $k$ for which the system has a unique solution.

If the system of equations $kx + 2y - z = 2, (k - 1)x + ky + z = 1, x + (k - 1)y + kz = 3$ has only one solution,then the number of possible real value$(s)$ of $k$ is -

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