For the system of linear equations:x - 2y = 1 | vedclass

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(D) The system of equations is:$x - 2y + 0z = 1$$x - y + kz = -2$$0x + ky + 4z = 6$The determinant of the coefficient matrix $D$ is:$D = \begin{vmatri

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$(A)$ and $(D)$ only

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(D) The system of equations is:$x - 2y + 0z = 1$$x - y + kz = -2$$0x + ky + 4z = 6$The determinant of the coefficient matrix $D$ is:$D = \begin{vmatrix} 1 & -2 & 0 \ 1 & -1 & k \ 0 & k & 4 \end{vmatrix} = 1(-4 - k^2) - (-2)(4 - 0) + 0 = -4 - k^2 + 8 = 4 - k^2$.For a unique solution,$D 
eq 0$,which implies $4 - k^2 
eq 0$,so $k 
eq 2$ and $k 
eq -2$. Thus,statement $(A)$ is correct.If $k = 2$,$D = 0$. We check for consistency using $D_1$:$D_1 = \begin{vmatrix} 1 & -2 & 0 \ -2 & -1 & 2 \ 6 & 2 & 4 \end{vmatrix} = 1(-4 - 4) - (-2)(-8 - 12) + 0 = -8 - 40 = -48 
eq 0$.Since $D = 0$ and $D_1 
eq 0$,the system has no solution when $k = 2$. Thus,statement $(D)$ is correct.Therefore,statements $(A)$ and $(D)$ are correct.

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