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Determinant of the following $2018 \times 2018$ matrix and let $B$ be the leading principal minor of $A$ of order $1009$, then rank of $B$

$$\begin{pmatrix} 0 & 2 & 0 & \ldots & \ldots & 0 \\ \frac{1}{3} & 0 & 2 & \ddots & & \vdots \\ 0 & \frac{1}{3} & 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ \vdots & & \ddots & \ddots & 0 & 2 \\ 0 & \ldots & \ldots & 0 & \frac{1}{3} & 0 \\ \end{pmatrix}.$$

I tried to find out some non- zero eigenvector to get the eigenvalues but I did'nt get it. Please help on this problem.

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  • $\begingroup$ So the diagonal is zero, the upper band is 2 and the lower band is 1/3? $\endgroup$ – ja72 Mar 13 '18 at 4:59
  • $\begingroup$ @ja72, yes ..... $\endgroup$ – 1256 Mar 13 '18 at 5:58
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For an $n$ dimensional matrix of the form

$$ A_n =\begin{pmatrix} 0 & u & 0 & & \cdots & 0 \\ l & 0 & u & & \cdots & \\ 0 & l & 0 \\ \vdots & & & \ddots & & \vdots \\ 0 & \cdots & & & 0 & u \\ 0 & \cdots & & & l & 0 \end{pmatrix} $$

The determinant is

$$ {\rm det} [A_n] = \begin{cases} 0 & n=\mbox{odd} \\ (-l u)^{\frac{n}{2}} & n=\mbox{even} \end{cases} $$

I think in your case $n=2018$, $u=2$ and $l=\frac{1}{3}$

$$ {\rm det}[A_{2018}] = \left( -\frac{2}{3} \right)^{1009} $$


I arrived at the rule by looking at the determinant of $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ using a CAS system.

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  • $\begingroup$ +1 nice answer, i like its $\endgroup$ – user476275 Apr 30 '18 at 7:03
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I'm not sure what you're precisely asking, but if it's just for the determinant of the matrix given, here's a hint: determinants are invariant under taking linear combinations of rows/columns. That is, subtracting a scalar multiple of a row from another doesn't change the determinant, and similarly for columns.

You need to perform some row swaps after that - again, the effect on the determinant is well-known.

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