If the system of equations x+y+z=1, x+2y+4z=k | vedclass

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(C) For the system of linear equations to be consistent, the determinant of the coefficient matrix $\Delta$ must be zero, and the determinants $\Delta

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$1, 2$

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(C) For the system of linear equations to be consistent, the determinant of the coefficient matrix $\Delta$ must be zero, and the determinants $\Delta_1, \Delta_2, \Delta_3$ must also be zero. $\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 4 \ 1 & 4 & 10 \end{vmatrix} = 1(20-16) - 1(10-4) + 1(4-2) = 4 - 6 + 2 = 0$. Since $\Delta = 0$, the system is consistent if $\Delta_1 = \Delta_2 = \Delta_3 = 0$. $\Delta_1 = \begin{vmatrix} 1 & 1 & 1 \ k & 2 & 4 \ k^2 & 4 & 10 \end{vmatrix} = 1(20-16) - 1(10k-4k^2) + 1(4k-2k^2) = 4 - 10k + 4k^2 + 4k - 2k^2 = 2k^2 - 6k + 4 = 2(k^2 - 3k + 2) = 2(k-1)(k-2)$. Setting $\Delta_1 = 0$, we get $(k-1)(k-2) = 0$, which implies $k = 1$ or $k = 2$. Similarly, $\Delta_2 = \begin{vmatrix} 1 & 1 & 1 \ 1 & k & 4 \ 1 & k^2 & 10 \end{vmatrix} = 1(10k-4k^2) - 1(10-4) + 1(k^2-k) = 10k - 4k^2 - 6 + k^2 - k = -3k^2 + 9k - 6 = -3(k^2 - 3k + 2) = -3(k-1)(k-2)$. Setting $\Delta_2 = 0$, we get $k = 1$ or $k = 2$. $\Delta_3 = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & k \ 1 & 4 & k^2 \end{vmatrix} = 1(2k^2-4k) - 1(k^2-k) + 1(4-2) = 2k^2 - 4k - k^2 + k + 2 = k^2 - 3k + 2 = (k-1)(k-2)$. Setting $\Delta_3 = 0$, we get $k = 1$ or $k = 2$. Thus, the system is consistent for $k = 1, 2$.

If the system of equations $x+y+z=1$, $x+2y+4z=k$ and $x+4y+10z=k^2$ is consistent, then $k$ is equal to

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