Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the system of linear equations  <img alt="x+3y+k^2z=0" src="http://entrancecorner.codecogs.com/gif.latex?x&plus;3y&plus;k%5E2z%3D0" /

If the system of linear equations  x+y+3z=0 x+3y+k^2z=0 3x+y+3z=0 has non zero solution (x,y,z)\: \: for\, \, \: \: some \: \: k\in R then x+\left ( \frac{y}{z} \right )=?   
Option: 1 -9
Option: 2 9
Option: 3 3
Option: 4 -3

Answers (1)

best_answer

For non-zero solutions to exist,

\\\begin{vmatrix} 1 &1 &3 \\ 1& 3 &k^2 \\ 3& 1 &3 \end{vmatrix}=0\\\Rightarrow 1(9-k^2)-1(3-3k^2)+3(1-9)=0\\k^2=9\\

Now the system of equation become

x + y + 3z = 0  

x + 3y + 9z = 0

3x + y + 3z = 0

we, get x = 0

3y = -9z

y/z = -3

x+ (y/z) = -3

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in
anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

Latest Question