Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

explain solution rd sharma class 12 chapter Solution of simultaneous linear equation exercise 7.2 question 4 maths

Answers (1)

Answer:

$x= 2k,y= 4k,z= k$

Hint:

If $\left | A \right |= 0$ , then the system of equation has non-trivial solution.

Given:

$x+y-6z=0$

$x-y+2z=0$

$-3x+y+2z=0$

Explanation:

The given homogeneous system can be written in matrix form

i.e. $\begin{bmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0 \end{bmatrix}$

$A=\begin{bmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{bmatrix}$

let

$\left | A \right |=\begin{vmatrix} 1 & 1 & -6\\ 1& -1 & 2\\ -3& 1 & 2 \end{vmatrix}$

$= 1(-2-2)-1(2+6)-6(1-3)$

$= -4-8+12$

$= 0$

So, the given system of equation has a non-trivial solution.

To find these solutions, we write the first two equations as

$x+y=6z$

$x-y=-2z$

Let $z=k$ , then we have

$\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 6k\\ -2k\end{bmatrix}$

$A=\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix},X=\begin{bmatrix} x\\ y\end{bmatrix},B=\begin{bmatrix} 6k\\ -2k\end{bmatrix}$

Where,

$\left | A \right |=\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}=-2\neq 0$

So $A^{-1}$ exists

^Now,

Put in third equation, so these valuessatisfies it.

Hence ^,, where

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon

anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers

anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question