Given,
(++3)x+(3−3+)y+(−4+5)z =λ(x+y+z) On equating the coefficients of
,and both sides, we have
x+3y−4z=λx, x−3y+5z=λy and
3x+y+0=λz Above three equations can b e rewritten as
(1−λ)x+3y−4z=0 x−(3+λ)y+5z=0 3x+y−λz=0 This is homogeneous systemn of equations in three variables x, y and z.
It is consistent and have non - zero solution .
i.e (x, y, z) ≠ (0, 0, 0), if determinant of coefficient matrix is zero.
⇒
[] On expanding along first row, we have
(1−λ)[λ(3+λ)−5]−3(−λ−15)−4(1+9+3λ)=0 ⇒
(1−λ)(λ2+3λ−5)+3λ+45−40−12λ=0 ⇒
λ2+3λ−5−λ3−3λ2+5λ−9λ+5=0 ⇒
−λ3−2λ2−λ=0 ⇒
λ(λ2+2λ+1)=0 ⇒
λ(λ+1)2=0 ⇒
λ=0,−1