1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# If the system of equation 2x+3yâˆ’z=0,x+kyâˆ’2z=0 and 2xâˆ’y+z=0 has a non trivial solution (x,y,z) then xy+yz+zx+k is equal to:

A
34
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
4
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
14
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C 12Given 2x+3y−z=0,x+Ky−2z=0,2x−y+z=0for Non-trivial solution∣∣ ∣∣23−11K−22−11∣∣ ∣∣=0By expanding along R2 ⇒−1[3−1]+K[2+2]−(−2)[−2−6]=0∴K=92the given equations become 2x+3y−z=0......(1)x+92y−2z=0.......(2)2x−y+z=0..........(3)(1)−(3)⇒4y−2z=02y=z........(4)yz=12.......(5)put z from equation (4) into (1)2x+3y−2y=02x+y=0xy=−12.....(6)(6)×(5)⇒xz=−14⇒zx=−4xy+yz+zx+k=12−12−4+92∴xy+yz+zx+k=12

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Properties
MATHEMATICS
Watch in App
Join BYJU'S Learning Program