If the system of equations2x + y - z = 52x - | vedclass

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(A) For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero:$\Delta = \begin{vmatrix} 2 &

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$916$

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(A) For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero:$\Delta = \begin{vmatrix} 2 & 1 & -1 \ 2 & -5 & \lambda \ 1 & 2 & -5 \end{vmatrix} = 0$Expanding along the first row:$2(25 - 2\lambda) - 1(-10 - \lambda) - 1(4 + 5) = 0$$50 - 4\lambda + 10 + \lambda - 9 = 0$$51 - 3\lambda = 0 \Rightarrow 3\lambda = 51 \Rightarrow \lambda = 17$For infinitely many solutions,the determinant $\Delta_x$ (or $\Delta_y$ or $\Delta_z$) must also be zero:$\Delta_z = \begin{vmatrix} 2 & 1 & 5 \ 2 & -5 & \mu \ 1 & 2 & 7 \end{vmatrix} = 0$Expanding along the first row:$2(-35 - 2\mu) - 1(14 - \mu) + 5(4 + 5) = 0$$-70 - 4\mu - 14 + \mu + 45 = 0$$-3\mu - 39 = 0 \Rightarrow 3\mu = -39 \Rightarrow \mu = -13$Now,calculate $(\lambda + \mu)^2 + (\lambda - \mu)^2 = 2(\lambda^2 + \mu^2)$:$= 2(17^2 + (-13)^2) = 2(289 + 169) = 2(458) = 916$

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

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