If the system of equations2x + 7y + lambda z | vedclass

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(A) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ and the determinants $D_1, D_2, D

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$38$

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(A) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ and the determinants $D_1, D_2, D_3$ must all be zero.First,we calculate $D_3 = 0$:$D_3 = \begin{vmatrix} 2 & 7 & 3 \ 3 & 2 & 4 \ 1 & \mu & -1 \end{vmatrix} = 2(-2 - 4\mu) - 7(-3 - 4) + 3(3\mu - 2) = 0$$-4 - 8\mu + 49 + 9\mu - 6 = 0$$\mu + 39 = 0 \Rightarrow \mu = -39$Next,we calculate $D = 0$ with $\mu = -39$:$D = \begin{vmatrix} 2 & 7 & \lambda \ 3 & 2 & 5 \ 1 & -39 & 32 \end{vmatrix} = 2(64 + 195) - 7(96 - 5) + \lambda(-117 - 2) = 0$$2(259) - 7(91) - 119\lambda = 0$$518 - 637 - 119\lambda = 0$$-119 - 119\lambda = 0 \Rightarrow \lambda = -1$Finally,we find $(\lambda - \mu)$:$\lambda - \mu = -1 - (-39) = -1 + 39 = 38$.

If the system of equations
$2x + 7y + \lambda z = 3$
$3x + 2y + 5z = 4$
$x + \mu y + 32z = -1$
has infinitely many solutions,then $(\lambda - \mu)$ is equal to

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If the system of equations$2x + 7y + \lambda z = 3$$3x + 2y + 5z = 4$$x + \mu y + 32z = -1$has infinitely many solutions,then $(\lambda - \mu)$ is equal to