If the system of equations 2x + lambda y + 3z | vedclass

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

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(C) 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 $0$.$\Delta = \begin{vmatrix} 2 & \lambda & 3 \ 3 & 2 & -1 \ 4 & 5 & \mu \end{vmatrix} = 2(2\mu + 5) - \lambda(3\mu + 4) + 3(15 - 8) = 0$$4\mu + 10 - 3\lambda\mu - 4\lambda + 21 = 0 \Rightarrow 4\mu - 3\lambda\mu - 4\lambda + 31 = 0 \dots (1)$Now,consider $\Delta_z = \begin{vmatrix} 2 & \lambda & 5 \ 3 & 2 & 7 \ 4 & 5 & 9 \end{vmatrix} = 0$$2(18 - 35) - \lambda(27 - 28) + 5(15 - 8) = 0$$2(-17) - \lambda(-1) + 5(7) = 0 \Rightarrow -34 + \lambda + 35 = 0 \Rightarrow \lambda = -1$Substituting $\lambda = -1$ into equation $(1)$:$4\mu - 3(-1)\mu - 4(-1) + 31 = 0$$4\mu + 3\mu + 4 + 31 = 0 \Rightarrow 7\mu = -35 \Rightarrow \mu = -5$Therefore,$\lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26$.

If the system of equations $2x + \lambda y + 3z = 5$,$3x + 2y - z = 7$,and $4x + 5y + \mu z = 9$ has infinitely many solutions,then $(\lambda^2 + \mu^2)$ is equal to:

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