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If $\alpha$ and $\beta$ are non real numbers satisfying $x^3 -1 = 0$ , then evaluate, \begin{vmatrix} \lambda +1 &\alpha &\beta\\ \alpha &\lambda + \beta &1\\ \beta &1 &\lambda + \alpha \end{vmatrix}

I tried this: $\alpha $ =$\omega$, $\beta = \omega^2$. Then I modified columns as $C_1 = C_1+C_2+C_3$ and further $R_1=R_1-R_3$ and $R_2=R_2-R_3$. to get this: \begin{vmatrix} 0 &\omega-1 &\omega ^2-\lambda-\omega\\ 0 & \omega^2+\lambda+1 &\lambda+\omega+1\\ \omega &1 &\lambda + \omega \end{vmatrix} I further simplified it to get the value of determinant as $\lambda[\lambda^2-1-\omega]$ .The answer is only $\lambda^3.$ Please point out any errors if you find them.

Shreshta S
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  • If alpha and beta are non real, then they are complex. Since $x=1$ satisfies the equation $x^3-1=0$ as a real number, the greek letters must satisfiy $x^2+x+1=0$ which can be calculated in different ways. In other words, alpha and beta are known – imranfat Aug 02 '16 at 17:08

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Using your notation but with $\;t=\lambda\;$ :$${}$$

$$\begin{vmatrix} t + 1&\alpha &\beta\\ \alpha &t + \beta &1\\ \beta &1 &t + \alpha \end{vmatrix}=\begin{vmatrix} t + 1&w &w^2\\ w &t +w^2 &1\\ w^2 &1 &t +w \end{vmatrix}=$$$${}$$

$$=(t+1)(t^2-t+1)+1+1-(t+1)-w^2(t+w)-w(t+w^2)=$$

$$=t^3+1+2-t-1-w^2t-1-wt-1=t^3-(w^2+w+1)t=t^3$$

Observe that $\;\alpha+\beta+1=w+w^2+1=0\;$ and from here the above relations.

DonAntonio
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  • Ah, you did not even calculate the roots of the given equation, much smarter... – imranfat Aug 02 '16 at 17:17
  • @imranfat That's trivial: they are the primitive roots of unit of order three. – DonAntonio Aug 02 '16 at 17:19
  • I'm sorry, I didn,t understand. Could you add a bit more details before the 2nd step? – Shreshta S Aug 03 '16 at 01:55
  • @ShreshtaS By your definition, $;\alpha= w=e^{2\pi i/3},,,,\beta=w^2=e^{4\pi i/3};$ are the two non-real, complex roots of $;x^3-1=(x-1)(x^2+x+1)\implies;$ they are roots of the second factor, and thus $;w^2+w+1=0;$ . The second line is the outcome of the determinant of the matrix in the first line, where I wrote your first determinant but with $;w,w^2;$ instead of $;\alpha,\beta;$ . – DonAntonio Aug 03 '16 at 07:09
  • no I meant the one after he determinant when substituted with $\alpha=\omega$ and $\beta=\omega^2$ – Shreshta S Aug 03 '16 at 16:55
  • You mean the second line of calculations? That's the determinant by diagonals. If you haven't seen that then reduce first somehow the matrix and evaluate. – DonAntonio Aug 03 '16 at 17:13