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For $n \geq 2,$ consider the following square matrix of order $(n-1)$

\begin{array}{cccccc} 3 & 1 & 1 & 1 & & 1 \\ 1 & 4 & 1 & 1 & \dots & 1 \\ 1 & 1 & 5 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 6 & & 1 \\ & & & & \ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & n+1 \end{array}

Find its determinant using only elementary row operations and denote it by $A_{n}$. Hence or otherwise, check whether the sequence $\left\{^{A_{n}} / n_{1}\right\}_{n>2}$ is bounded

Basically I am getting no clue. Should I proceed with Echelon Form?

1 Answers1

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Basic idea:
1. We get rid of $1$s except for the first row
2. We get rid of $1$s in the first row
3. We consider $(A)_{1,1}$ as the rest is known

So 1. we subtract the 1st row from every other row: $$\begin{align*} \left|\begin{array}{cccccc} 3 & 1 & 1 & 1 & & 1 \\ 1 & 4 & 1 & 1 & \dots & 1 \\ 1 & 1 & 5 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 6 & & 1 \\ & & & & \ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & n+1 \end{array}\right| &= \left|\begin{array}{cccccc} 3 & 1 & 1 & 1 & & 1 \\ -2 & 3 & 0 & 0 & \dots & 0 \\ -2 & 0 & 4 & 0 & \cdots & 0 \\ -2 & 0 & 0 & 5 & & 0 \\ & & & & \ddots & \vdots \\ -2 & 0 & 0 & 0 & \cdots & n \end{array}\right| \end{align*}$$ 2. we subtract each row (say $k$th), multiplied by $\frac{1}{k+1}$ from the first row: $$= \left|\begin{array}{cccccc} a_{1,1} & 0 & 0 & 0 & & 0 \\ -2 & 3 & 0 & 0 & \dots & 0 \\ -2 & 0 & 4 & 0 & \cdots & 0 \\ -2 & 0 & 0 & 5 & & 0 \\ & & & & \ddots & \vdots \\ -2 & 0 & 0 & 0 & \cdots & n \end{array}\right|$$ 3. As it's lower-triangular now, we consider $a_{1,1}$ as $|A|=a_{1,1}\prod\limits_{k=3}^n k$,
$a_{1,1}=3+2\sum\limits_{k=3}^n \frac{1}{k}$

Alexey Burdin
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  • It would be better if you throughly explain! –  May 24 '20 at 20:05
  • Please see the edits. Is the part about sequence needed to be written out or it's obvious? – Alexey Burdin May 24 '20 at 20:35
  • determinant of lower triangular matrix is a well-known result, as for an upper triangular too. On $a_{1,1}$: It was $3$ on step 1. Then we subtracted from it $-2$ for each oher row, multiplied by $\frac{1}{k+1}$ for $k$ from $2$ to $n-1$ -- the same thing as added $2\cdot\frac{1}{k}$ for $k$ from $3$ to $n$. Sure it was step 3, not 2? Or the prod/sum notations? – Alexey Burdin May 24 '20 at 21:12
  • sorry! step 3 is clear to me! I am unable to understand about Step 2, why we subtract each row (say kth), multiplied by 1k+1 from the first row and how is it working? –  May 24 '20 at 21:15
  • Is the legality of such operation needed to be shown somehow? Why: to get rid of the $1$s in the first row. How it's working: on $k$th row (for $k>1$) now on the step 2 (before we proceed) there are only $2$ elements $\ne 0$, namely $a_{k,1}=-2$ and $a_{k,k}=k+1$, so we want $a_{k,k}$ to eliminate $a_{1,k}$ therefore the constant is chosen to be $\frac{1}{k+1}$ so that $a_{1,k}-\frac{1}{k+1}a_{k,k}=1-\frac{1}{k+1}\cdot(k+1)=0$ – Alexey Burdin May 24 '20 at 21:27
  • great.now it is cleared! –  May 24 '20 at 21:28
  • Let us continue this discussion in chat. –  May 24 '20 at 21:33