If A=[1ωω2ωω21ω21ω],B=[ωω21ω21ωωω21] and C=[1ωω2] where ω is the complex cube root of 1 then (A+B)C is equal to

A+B=[1+ωω+ω2ω2+1ω+ω2ω2+11+ωω2+ω1+ω2ω+1]=[−ω2−1−ω−1−ω−ω2−1−ω−ω2] ∴(A+B)C=[−ω2−1−ω−1−ω−ω2−1−ω−ω2][1ωω2]=[−ω2−ω−ω3−1−ω2−ω4−1−ω2−ω]=[000]using ω3=1,1+ω+ω2=0

If A=[1ωω2ωω21ω21ω],B=[ωω21ω21ωωω21] and C=[1ωω2] where ω is the complex cube root of 1 then (A+B)C is equal to

A+B=[1+ωω+ω2ω2+1ω+ω2ω2+11+ωω2+ω1+ω2ω+1]=[−ω2−1−ω−1−ω−ω2−1−ω−ω2] ∴(A+B)C=[−ω2−1−ω−1−ω−ω2−1−ω−ω2][1ωω2]=[−ω2−ω−ω3−1−ω2−ω4−1−ω2−ω]=[000]using ω3=1,1+ω+ω2=0

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If $A = [\begin{matrix} 1 & ω & ω^{2} \\ ω & ω^{2} & 1 \\ ω^{2} & 1 & ω \end{matrix}], B = [\begin{matrix} ω & ω^{2} & 1 \\ ω^{2} & 1 & ω \\ ω & ω^{2} & 1 \end{matrix}]$ and $C = [\begin{array}{l} 1 \\ ω \\ ω^{2} \end{array}]$ where $ω$ is the complex cube root of 1 then $(A + B) C$ is equal to

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$[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}]$

$[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$

answer is A.

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Detailed Solution

$A + B = [\begin{matrix} 1 + ω & ω + ω^{2} & ω^{2} + 1 \\ ω + ω^{2} & ω^{2} + 1 & 1 + ω \\ ω^{2} + ω & 1 + ω^{2} & ω + 1 \end{matrix}] = [\begin{matrix} - ω^{2} & - 1 & - ω \\ - 1 & - ω & - ω^{2} \\ - 1 & - ω & - ω^{2} \end{matrix}] ∴ (A + B) C = [\begin{matrix} - ω^{2} & - 1 & - ω \\ - 1 & - ω & - ω^{2} \\ - 1 & - ω & - ω^{2} \end{matrix}] [\begin{array}{l} 1 \\ ω \\ ω^{2} \end{array}] = [\begin{array}{l} - ω^{2} - ω - ω^{3} \\ - 1 - ω^{2} - ω^{4} \\ - 1 - ω^{2} - ω \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$