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I need help with this problem:

For every $i,j \in \{1,2,...,n\}$ is $d_{i,j}=min\{i,j\}$. Calculate determinant of a matrix $[d_{i,j}]_{n_Xn}$.

Is it right that all the elements of this squared matrix are $1$, because of that $d_{i,j}=min\{i,j\}$? Then, value of determinant is $0$.

Thanks for replies.

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    $\begingroup$ Why don't you just try a $2 \times 2$ example? What is $d_{2,2}$? $\endgroup$ – copper.hat May 20 '15 at 18:18
  • $\begingroup$ @ copper.hat But first I need to form a matrix. Is it right that all elements are $1$ by$d_{i,j}=min\{i,j\}$? $\endgroup$ – user300045 May 20 '15 at 18:20
  • $\begingroup$ I don't understand what you are asking. There is a formula for $d_{i,j}$ which gives the $i,j$ entry. $\endgroup$ – copper.hat May 20 '15 at 18:21
  • $\begingroup$ @ copper.hat I am asking what are the elements of a matrix? And what is that formula? $\endgroup$ – user300045 May 20 '15 at 18:23
  • $\begingroup$ The formula you gave above??? The $i,j$ entry is $\min(i,j)$. $\endgroup$ – copper.hat May 20 '15 at 18:24
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This matrix has determinant $1$, see here. The matrix looks as follows $$ \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \cr 1 & 2 & 2 & \cdots & 2 \cr 1 & 2 & 3 &\cdots & 3 \cr \vdots & \vdots & \vdots & \ddots & \vdots \cr 1 & 2 & 3 & \cdots & n \end{pmatrix} $$

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  • $\begingroup$ How to form that matrix if we only know the statement $d_{i,j}=min\{i,j\}$? $\endgroup$ – user300045 May 20 '15 at 18:28
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    $\begingroup$ For example, $d_{2,3}=min(2,3)=2$, so the coefficient at place $(2,3)$ is equal to $2$. $\endgroup$ – Dietrich Burde May 20 '15 at 18:29

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