If the system of equations x - ky - z = 0,kx | vedclass

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(D) For a system of linear equations to have a non-zero (non-trivial) solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\

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$-1, 1$

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(D) For a system of linear equations to have a non-zero (non-trivial) solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\Delta = 0$.The system is given by:$x - ky - z = 0$$kx - y - z = 0$$x + y - z = 0$The determinant $\Delta$ is:$\Delta = \begin{vmatrix} 1 & -k & -1 \ k & -1 & -1 \ 1 & 1 & -1 \end{vmatrix} = 0$Expanding along the first row:$1((-1)(-1) - (-1)(1)) - (-k)((k)(-1) - (-1)(1)) + (-1)((k)(1) - (-1)(1)) = 0$$1(1 + 1) + k(-k + 1) - 1(k + 1) = 0$$2 - k^2 + k - k - 1 = 0$$1 - k^2 = 0$$k^2 = 1$$k = \pm 1$Thus,the possible values of $k$ are $1$ and $-1$.

If the system of equations $x - ky - z = 0$,$kx - y - z = 0$ and $x + y - z = 0$ has a non-zero solution,then the possible values of $k$ are:

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