Is the answer also imaginary?

Algebra Level 4

If ω \omega is one of the imaginary cube roots of unity,evaluate ∣ 1 ω ω 2 ω 3 ω ω 2 ω 3 1 ω 2 ω 3 1 ω ω 3 1 ω ω 2 ∣ 2 . \left | \begin{array}{cccc} 1 & \omega & \omega^2 & \omega^3 \\ \omega & \omega^2 & \omega^3 & 1 \\ \omega^2 & \omega^3 & 1 & \omega \\ \omega^3 & 1 & \omega & \omega^2 \end{array} \right |^{2}. If the value of above determinant is in the form a + b − 1 a+b\sqrt{-1} , find a + b a+b .


The answer is -27.

Rohit Udaiwal by Rohit Udaiwal , 20, India

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1 solution

Rishabh Jain
Jan 8, 2016

∣ 1 ω ω 2 1 1 ω 2 1 1 1 1 1 ω 1 1 ω ω 2 ∣ 2 \left | \begin{array}{cccc} 1 & \omega & \omega^2 & 1 \\ 1 & \omega^2 & 1 & 1 \\ 1 & 1 & 1 & \omega \\ 1 & 1 & \omega & \omega^2 \end{array} \right |^{2} U s i n g ω 2 + ω + 1 = 0 , C 1 → C 1 + C 2 + C 3 + C 4 Using\space \omega^2+\omega+1=0 \space \color{#D61F06}{, C_1\rightarrow C_1+C_2+C_3+C_4} ∣ 1 ω ω 2 1 0 − ω ( 1 − ω ) − ω 2 ( 1 − ω ) 0 0 ( 1 − ω ) − ω 2 ( 1 − ω ) − ( 1 − ω ) 0 ( 1 − ω ) ω ( 1 − ω ) ω 2 ( 1 − ω ) ∣ 2 \left | \begin{array}{cccc} 1 & \omega & \omega^2 & 1 \\ 0 & -\omega(1-\omega) & -\omega^2(1-\omega) & 0 \\ 0 & (1-\omega) & -\omega^2(1-\omega) & -(1-\omega) \\ 0 & (1-\omega) & \omega(1-\omega) & \omega^2(1-\omega)\end{array} \right |^{2} R 2 → R 2 − R 1 , R 3 → R 3 − R 1 , R 4 → R 4 − R 1 \color{#D61F06}{R_2 \rightarrow R_2-R_1,R_3\rightarrow R_3-R_1,R_4 \rightarrow R_4-R_1} = ( 1 − ω ) 6 ∣ 1 ω ω 2 1 0 − ω ω 2 0 0 − 1 − ω 2 − 1 0 1 ω − ω 2 ∣ 2 =(1-\omega)^6 \left | \begin{array}{cccc} 1 & \omega & \omega^2 & 1 \\ 0 & -\omega & \omega^2 & 0 \\ 0 & -1 & -\omega^2 & -1 \\ 0 & 1 & \omega & -\omega^2 \end{array} \right |^{2} = ( 1 − ω ) 6 = − 27 =(1-\omega)^6\color{magenta}{=-27}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you for such a nice solution!

Rohit Udaiwal - 5 years, 5 months ago

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