If the system of equations 2x - y + z = 4,5x | vedclass

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determin

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$57$

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\Delta_x, \Delta_y, \Delta_z$ must also be $0$.First,we calculate $\Delta = \begin{vmatrix} 2 & -1 & 1 \ 5 & \lambda & 3 \ 100 & -47 & \mu \end{vmatrix} = 2(\lambda\mu + 141) + 1(5\mu - 300) + 1(-235 - 100\lambda) = 0$.Simplifying,we get $2\lambda\mu + 282 + 5\mu - 300 - 235 - 100\lambda = 0$,which gives $2\lambda\mu + 5\mu - 100\lambda = 253$ (Equation $1$).Next,we calculate $\Delta_z = \begin{vmatrix} 2 & -1 & 4 \ 5 & \lambda & 12 \ 100 & -47 & 212 \end{vmatrix} = 0$.Expanding along the first row: $2(212\lambda + 564) + 1(1060 - 1200) + 4(-235 - 100\lambda) = 0$.$424\lambda + 1128 - 140 - 940 - 400\lambda = 0$.$24\lambda + 48 = 0 \Rightarrow \lambda = -2$.Substituting $\lambda = -2$ into Equation $1$: $2(-2)\mu + 5\mu - 100(-2) = 253$.$-4\mu + 5\mu + 200 = 253 \Rightarrow \mu = 53$.Finally,we calculate $\mu - 2\lambda = 53 - 2(-2) = 53 + 4 = 57$.

If the system of equations $2x - y + z = 4$,$5x + \lambda y + 3z = 12$,and $100x - 47y + \mu z = 212$ has infinitely many solutions,then $\mu - 2\lambda$ is equal to

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