If the system of linear equations 8x + y + 4z | vedclass

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(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$.$D = \begin{vmatrix

Vedclass · Accepted Answer

$\frac{10}{3}$

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(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$.$D = \begin{vmatrix} 8 & 1 & 4 \ 1 & 1 & 1 \ \lambda & -3 & 0 \end{vmatrix} = 8(0 - (-3)) - 1(0 - \lambda) + 4(-3 - \lambda) = 0$$24 + \lambda - 12 - 4\lambda = 0 \Rightarrow 12 - 3\lambda = 0 \Rightarrow \lambda = 4$.Now,for infinitely many solutions,the augmented matrix must be consistent with rank less than $3$. Using $D_1 = 0$:$D_1 = \begin{vmatrix} -2 & 1 & 4 \ 0 & 1 & 1 \ \mu & -3 & 0 \end{vmatrix} = -2(0 - (-3)) - 1(0 - \mu) + 4(0 - \mu) = 0$$-6 + \mu - 4\mu = 0 \Rightarrow -3\mu = 6 \Rightarrow \mu = -2$.The point is $\left(4, -2, -\frac{1}{2}ight)$.The distance of the point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $\frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Distance $= \frac{|8(4) + 1(-2) + 4(-\frac{1}{2}) + 2|}{\sqrt{8^2 + 1^2 + 4^2}} = \frac{|32 - 2 - 2 + 2|}{\sqrt{64 + 1 + 16}} = \frac{30}{\sqrt{81}} = \frac{30}{9} = \frac{10}{3}$.

If the system of linear equations $8x + y + 4z = -2$,$x + y + z = 0$,and $\lambda x - 3y = \mu$ has infinitely many solutions,then the distance of the point $\left(\lambda, \mu, -\frac{1}{2}\right)$ from the plane $8x + y + 4z + 2 = 0$ is:

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