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Calculate the determinant of the following matrix as an explicit function of $x$. (It is a polynomial in $x$. You are asked to find all the coefficients.)

\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\ x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\ 0 & 0 & 0 & x^{10} & x^{11}\\ 0 & 0 & 0 & x^{12} & x^{13}\\ 0 & 0 & 0 & x^{14} & x^{15} \end{bmatrix}

Can someone help me with this question?

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    $\begingroup$ matrix blocks $ $ $\endgroup$ – leo Oct 27 '13 at 23:30
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Hint: $$\det\begin{pmatrix}A & 0\\ C & D\end{pmatrix} = \det\begin{pmatrix}A & B\\ 0 & D\end{pmatrix} = \det(A) \det(D)$$ Where $A,B,C,D$ are block matrices.

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  • $\begingroup$ should I choose A as upperleft two rows and columns and D as lower right 3 rows and columns? $\endgroup$ – Mark Oct 27 '13 at 23:39
  • $\begingroup$ wait, I didn't even have to do anything. A that I chose has determinant of zero so the whole matrix will have determinant of zero right? $\endgroup$ – Mark Oct 27 '13 at 23:41
  • $\begingroup$ A is any $n$ by $n$ block on the top left. You can indeed choose $n=3$ and get zero as you stated. $\endgroup$ – nbubis Oct 27 '13 at 23:48
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First, note that the 5th column is a multiple of the 4th column. That is,

\begin{bmatrix} x^4\\ x^9\\ x^{11}\\ x^{13}\\ x^{15}\\ \end{bmatrix}

is $x$ times \begin{bmatrix} x^3\\ x^8\\ x^{10}\\ x^{12}\\ x^{14}\\ \end{bmatrix}.

Because the determinant of a matrix does not change when you subtract a multiple of one column from another column, we get that that matrix has the same determinant as that of

\begin{bmatrix} 1 & x & x^2 & x^3 & 0\\ x^5 & x^6 & x^7 & x^8 & 0\\ 0 & 0 & 0 & x^{10} & 0\\ 0 & 0 & 0 & x^{12} & 0\\ 0 & 0 & 0 & x^{14} & 0\\ \end{bmatrix}

and you can easily tell that the determinant of that matrix is $0$.

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  • $\begingroup$ Now I see. This is a another great way to approach determinant. Thanks for the beautiful insight $\endgroup$ – Mark Oct 27 '13 at 23:42
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    $\begingroup$ Or, the determinant of a matrix in which its columns are linearly dependent is 0. $\endgroup$ – leo Oct 27 '13 at 23:42
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Another way to look at this: the bottom three rows can't have rank more than $2$, since they have only two nonzero columns, so the whole matrix can't have rank more than $4$, and therefore is singular.

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