Question

Q2. Find the value of a for which the system of equations has non - trivial
solution.
2(sin a) x+y
– 2z = 0.
3x – 2(Cos 2 a) y - 3z= 0,
5x – 3y - z=0​

Answer 1

For non-trivial solution D=0

or

∣

∣

∣

∣

∣

∣

∣

∣

1

3

2

k

k

3

3

−2

−4

∣

∣

∣

∣

∣

∣

∣

∣

=0

Apply R

2

−3R

1

,R

3

−2R

1

∴△=

∣

∣

∣

∣

∣

∣

∣

∣

1

0

0

k

−2k

3−2k

3

−11

−10

∣

∣

∣

∣

∣

∣

∣

∣

=0

or 20k+11(3−2k)=0

or 33−2k=0

∴k=33/2

Putting the value of k, the equations are

x+

2

33

y+3z=0 .......(1)

3x+

2

33

y−2z=0 .......(2)

2x+3y−4z=0 ........(3)

Mutliply (1) by (3) and subtract from (2) and similarly multiply (1) by 2 and subtract from (3). Thus we get the equivalent system of equations as

x+

2

33

y+3z=0

−33y−11z=0

−30y−10z=0

From any of the last two, we get 3y=−z

or

1

y

=

−3

z

=λ, say.

∴y=λ,z=−3λ.

From (1), we get

x+

2

33

y+3z=0

x+

2

33

λ−9λ=0∴x=−

2

15

λ

∴x:y:z=−

2

15

:1:−3

Answer 2

Answer:

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