The system of equations -k x+3 y-14 z=25,-15 | vedclass

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(D) The system is consistent if the determinant $\Delta 
eq 0$ or if $\Delta = 0$ and the system has infinitely many solutions.First,calculate the de

Vedclass · Accepted Answer

$R -\{-11, 11\}$

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(D) The system is consistent if the determinant $\Delta 
eq 0$ or if $\Delta = 0$ and the system has infinitely many solutions.First,calculate the determinant $\Delta$ of the coefficient matrix:$\Delta = \begin{vmatrix} -k & 3 & -14 \ -15 & 4 & -k \ -4 & 1 & 3 \end{vmatrix} = -k(12 + k) - 3(-45 - 4k) - 14(-15 + 16) = -12k - k^2 + 135 + 12k - 14 = 121 - k^2$.For a unique solution,$\Delta 
eq 0$,which implies $121 - k^2 
eq 0$,so $k 
eq \pm 11$.If $k = 11$,$\Delta = 0$. We check for consistency using $\Delta_x, \Delta_y, \Delta_z$:$\Delta_z = \begin{vmatrix} -11 & 3 & 25 \ -15 & 4 & 3 \ -4 & 1 & 4 \end{vmatrix} = -11(16 - 3) - 3(-60 + 12) + 25(-15 + 16) = -11(13) - 3(-48) + 25(1) = -143 + 144 + 25 = 26 
eq 0$.Since $\Delta = 0$ and $\Delta_z 
eq 0$,the system is inconsistent for $k = 11$.If $k = -11$,$\Delta = 0$. We check $\Delta_z$:$\Delta_z = \begin{vmatrix} 11 & 3 & 25 \ -15 & 4 & 3 \ -4 & 1 & 4 \end{vmatrix} = 11(16 - 3) - 3(-60 + 12) + 25(-15 + 16) = 11(13) - 3(-48) + 25(1) = 143 + 144 + 25 = 312 
eq 0$.Since $\Delta = 0$ and $\Delta_z 
eq 0$,the system is inconsistent for $k = -11$.Thus,the system is consistent for all $k \in R - \{11, -11\}$.

The system of equations $-k x+3 y-14 z=25$,$-15 x+4 y-k z=3$,and $-4 x+y+3 z=4$ is consistent for all $k$ in the set

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