Let the system of equations x+5y-z=1,4x+3y-3z | vedclass

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(A) For the system to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$ and the augmented determinants $\D

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$3$

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(A) For the system to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$ and the augmented determinants $\Delta_1, \Delta_2, \Delta_3$ must also be $0$.$\Delta = \begin{vmatrix} 1 & 5 & -1 \ 4 & 3 & -3 \ 24 & 1 & \lambda \end{vmatrix} = 1(3\lambda + 3) - 5(4\lambda + 72) - 1(4 - 72) = 3\lambda + 3 - 20\lambda - 360 + 68 = -17\lambda - 289 = 0$.Thus,$\lambda = -17$.Now,$\Delta_1 = \begin{vmatrix} 1 & 5 & -1 \ 7 & 3 & -3 \ \mu & 1 & -17 \end{vmatrix} = 1(-51 + 3) - 5(-119 + 3\mu) - 1(7 - 3\mu) = -48 + 595 - 15\mu - 7 + 3\mu = 540 - 12\mu = 0$.Thus,$\mu = 45$.The system becomes $x+5y-z=1$ and $4x+3y-3z=7$.Subtracting $3 	imes (Eq 1)$ from $(Eq 2)$: $(4x+3y-3z) - 3(x+5y-z) = 7 - 3(1) \Rightarrow x - 12y = 4 \Rightarrow x = 12y + 4$.Substituting $x$ into $Eq 1$: $(12y+4) + 5y - z = 1 \Rightarrow z = 17y + 3$.We are given $7 \leq x+y+z \leq 77$.Substituting $x$ and $z$: $7 \leq (12y+4) + y + (17y+3) \leq 77 \Rightarrow 7 \leq 30y + 7 \leq 77 \Rightarrow 0 \leq 30y \leq 70 \Rightarrow 0 \leq y \leq 2.33$.Since $y$ is an integer,$y \in \{0, 1, 2\}$.For each $y$,we get a unique integer solution for $x$ and $z$.Thus,there are $3$ solutions.

Let the system of equations $x+5y-z=1$,$4x+3y-3z=7$,$24x+y+\lambda z=\mu$,where $\lambda, \mu \in R$,have infinitely many solutions. Then the number of solutions of this system,if $x, y, z$ are integers and satisfy $7 \leq x+y+z \leq 77$,is

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