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Let $A$ be a $3\times3$ complex matrix and $B$ its transpose. Let $ a$ be a complex number such that $ a \neq1$ and $\det(A+a*B)=0$. Compute $\det(A+B)$ in terms of $a$ and $\det(A).$

I tried to use the polynomial expansion $ \det(A+xB)=\det A + q*x +w*x^2+ \det B*x^3 $ for any matrices $A,B$. Probably I should have found some relations between coefficients $q$ and $w $ beacuse $B$ is $A$ transposed, but got stuck.

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If you want to relate the two coefficients, note that $$ \det(A + xB) = \det([xA + B]^T) = \det(xA + B) = x^3 \det(A + x^{-1}B) $$ So, we have $$ \det A + q\,x +w\,x^2+ \det B\,x^3 = \\ x^{3}(\det A + q*x^{-1} +w*x^{-2}+ \det B*x^{-3}) =\\ \det B + w \,x + q\, x^2 + x^3\det A $$ So, we have $q = w$. Moreover, we may use the equation $\det(A + a\,B) = 0$ to say $$ (a^3 + 1) \det(A) + (a + a^2)q = 0 \implies\\ q = -\frac{a^3 + 1}{a + a^2} \det(A) $$ You should be able to finish it from there.

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  • $\begingroup$ It seems odd that the problem says $a \ne 1$; given your denominator, do you suppose it should have been $a \ne -1$? $\endgroup$ – John Hughes Mar 2 '15 at 13:59
  • $\begingroup$ @JohnHughes I do suppose exactly that. We could cancel the common factor, though. $\endgroup$ – Omnomnomnom Mar 2 '15 at 14:00

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