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I want to evaluate the determinant of the $n \times n$ matrix

$\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\ 0 & 0 & \ldots & 0 & -a\\ 0 & 0 & \ldots & -a & 0\\ &&&\vdots \\ 0 & -a & 0 &\dots & 0 \end{array}\right|.$

So I try to say that it is $(-1)^{ n + (n-1) + \ldots n-(n-2)}(-a)^{n-1}$. So power of -1 should be $\frac{(n-1)(n+2)}{2} + n-1$. However answer given is $n(n-1)/2$. Where is wrong?

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2 Answers 2

up vote 5 down vote accepted

In fact, your answer is correct and agrees with the given answer. Here is the reason:$$\frac{(n-1)(n+2)}{2} + n-1=\frac{n(n-1)+2(n-1)}{2}+n-1=\frac{n(n-1)}{2}+2(n-1),$$ which implies that $$(-1)^{\frac{(n-1)(n+2)}{2} + n-1}=(-1)^{\frac{n(n-1)}{2}+2(n-1)}=(-1)^{\frac{n(n-1)}{2}}.$$

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Hint for OP: $(-1)^{2(n-1)} = ((-1)^2)^{n-1} = 1^{n-1} = 1.$ –  user2468 Mar 4 '12 at 6:58

Using cofactor expansion, we have: $$ \left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\ 0 & 0 & \ldots & 0 & -a\\ 0 & 0 & \ldots & -a & 0\\ &&&\vdots \\ 0 & -a & 0 &\dots & 0 \end{array}\right| = \left|\begin{array}{ccccc} 0 & \ldots & 0 & -a\\ 0 & \ldots & -a & 0\\ &&&\vdots \\ -a & 0 &\dots & 0 \end{array}\right| . $$

The determinant of the anti-diagonal matrix in the RHS is give by (note that RHS is $n-1\times n-1$): $$ \det = \dfrac{(-1)^{\frac{n(n-1)}{2}} (-a)^n }{(-a)} = (-1)^{\frac{n(n-1)}{2}}(-a)^{n-1}.$$

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