If the system of linear equations x - 2y + kz | vedclass

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(B) For the system to have a non-zero solution,the determinant of the coefficient matrix must be zero:$\Delta = \begin{vmatrix} 1 & -2 & k \ 2 & 1 &

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$4x - 3y - 4 = 0$

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(B) For the system to have a non-zero solution,the determinant of the coefficient matrix must be zero:$\Delta = \begin{vmatrix} 1 & -2 & k \ 2 & 1 & 1 \ 3 & -1 & -k \end{vmatrix} = 0$Expanding along the first row:$1(-k + 1) - (-2)(-2k - 3) + k(-2 - 3) = 0$$-k + 1 + 2(-2k - 3) - 5k = 0$$-k + 1 - 4k - 6 - 5k = 0$$-10k - 5 = 0 \Rightarrow k = -\frac{1}{2}$Substituting $k = -\frac{1}{2}$ into the equations:$x - 2y - \frac{1}{2}z = 1 \Rightarrow 2x - 4y - z = 2$  $(1)$$2x + y + z = 2$  $(2)$Adding $(1)$ and $(2)$:$(2x - 4y - z) + (2x + y + z) = 2 + 2$$4x - 3y = 4$Thus,$(x, y)$ lies on the line $4x - 3y - 4 = 0$.

If the system of linear equations $x - 2y + kz = 1$,$2x + y + z = 2$,and $3x - y - kz = 3$ has a non-zero solution $(x, y, z) \neq 0$,then $(x, y)$ lies on the straight line whose equation is

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