If 1,ω ,ω 2 are the cube roots of unity and if [ beginmatrix 1+ω 2ω -2ω -b endmatrix ]+[ beginmatrix a -ω 3ω 2 endmatrix ]=[ beginmatrix 0 ω ω 1 endmatrix ], then a2+b2 is equal to

Question

If 1,ω ,ω 2 are the cube roots of unity and if [ beginmatrix 1+ω 2ω -2ω -b endmatrix ]+[ beginmatrix a -ω 3ω 2 endmatrix ]=[ beginmatrix 0 ω ω 1 endmatrix ], then a2+b2 is equal to (A)  1+ω 2  (B)  ω 2-1  (C)  1+ω  (D)  (1+ω )2

Tardigrade Admin · Accepted Answer

Given, [ beginmatrix 1+ω 2ω -2ω -b endmatrix ]+[ beginmatrix a -ω 3ω 2 endmatrix ]=[ beginmatrix 0 ω ω 1 endmatrix ] ⇒ [ beginmatrix 1+ω +a ω ω 2-b endmatrix ]=[ beginmatrix 0 ω ω 1 endmatrix ] ⇒ 1+ω +a=0,2-b=1 ⇒ a=-1-ω ,b=1 ∴ a2+b2=(-1-ω 2)+12 =1+ω 2+2ω +12 =0+ω +1 (∵ 1+ω +ω 2=0) =1+ω

Q. If $1, ω, ω^{2}$ are the cube roots of unity and if $[1 + ω - 2 ω 2 ω - b] + [a 3 ω - ω 2] = [0 ω ω 1],$ then $a^{2} + b^{2}$ is equal to

$1 + ω^{2}$

$ω^{2} - 1$

$1 + ω$

$(1 + ω)^{2}$

$ω^{2}$

Q. If 1,ω,ω2 are the cube roots of unity and if [1+ω−2ω​2ω−b​]+[a3ω​−ω2​]=[0ω​ω1​], then a2+b2 is equal to

1+ω2

ω2−1

1+ω

(1+ω)2

ω2