If the system of equations kx + (k+1)y + (k-1 | vedclass

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(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero. Let $D$ be the

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$-\frac{1}{2}$

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(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero. Let $D$ be the determinant of the coefficient matrix: $D = \begin{vmatrix} k & k+1 & k-1 \ k-1 & k+2 & k \ k+1 & k & k+2 \end{vmatrix} = 0$. Applying column operations $C_1 	o C_1 + C_2 + C_3$: $D = \begin{vmatrix} 3k & k+1 & k-1 \ 3k+1 & k+2 & k \ 3k+3 & k & k+2 \end{vmatrix} = 0$. This simplifies to $D = (3k+1) \begin{vmatrix} 1 & k+1 & k-1 \ 1 & k+2 & k \ 1 & k & k+2 \end{vmatrix} = 0$. Subtracting $R_1$ from $R_2$ and $R_3$: $D = (3k+1) \begin{vmatrix} 1 & k+1 & k-1 \ 0 & 1 & 1 \ 0 & -1 & 3 \end{vmatrix} = 0$. Expanding the determinant: $(3k+1) [1(3 - (-1))] = 0$. $(3k+1)(4) = 0$. $3k+1 = 0 \implies k = -\frac{1}{3}$. Wait,re-evaluating the determinant: $D = k((k+2)(k+2) - k^2) - (k+1)((k-1)(k+2) - k(k+1)) + (k-1)(k(k-1) - (k+2)(k+1)) = 0$. $D = k(k^2+4k+4-k^2) - (k+1)(k^2+k-2-k^2-k) + (k-1)(k^2-k-k^2-3k-2) = 0$. $D = k(4k+4) - (k+1)(-2) + (k-1)(-4k-2) = 0$. $4k^2+4k + 2k+2 - 4k^2-2k+4k+2 = 0$. $8k + 4 = 0 \implies 8k = -4 \implies k = -\frac{1}{2}$. Thus,the only possible value for $k$ is $-\frac{1}{2}$,and the sum of all possible values is $-\frac{1}{2}$.

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

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