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

JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

If the system of linear equations.

\begin{gathered} 8x + y + 4z = - 2 \quad \\ x + y + z = 0 \quad \\ \lambda x - 3y = \mu \quad \\ \end{gathered}

has infinitely many solutions, then the distance of the point

\left(\lambda, \mu,-\frac{1}{2}\right)

from the plane

8 x+y+4 z+2=0

View answer and explanation

Correct option: D

\frac{10}{3}

Option D (

\frac{10}{3}

) is correct because it represents the distance of the point from the plane after finding the values of

\lambda

and

\mu

for which the system has infinitely many solutions.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|