Determinant of the $n$-th order calculated with a combination of methods-verification

Question

$A\in M_n(\mathbb F)$ $$\det{A}=\begin{vmatrix} -1 &\;1&\;1&...&\;1&\;1&\;1 \\ -2 &-1&\;0&\ldots&\;0&\;0&\;1\\ -2&\;0&-1&...&\;0&\;0&\;1\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\-2&0&\;0&\ldots&-1&\;0&\;1\\-2&\;0&\;0&\ldots&\;0&-1&\;1\\-2&-2&-2&\ldots&-2&-2&-1 \end{vmatrix}=?$$

My work: I subtracted $\text{the last ($n$-th) row}$ so as to free the terrain for the LaPlace transform to the $\text{first column}$. I got: $$\begin{vmatrix} -1 &\;1&\;1&\ldots&\;1&\;1&\;1 \\ 0 &1&\;2&\ldots&\;2&\;2&\;2\\ 0&\;2&1&\ldots&\;2&\;2&\;2\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\0&2&\;2&\ldots&\;1&\;2&\;2\\\;0&\;2&\;2&\ldots&\;2&\;1&\;2\\-2&-2&-2&\ldots&-2&-2&-1 \end{vmatrix}$$ Then I have two sumands: $$-1\cdot\begin{vmatrix} \;1&\;2&\ldots&\;2&\;2&\;2\\ \;2&1&\ldots&\;2&\;2&\;2\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;2&\;2&\ldots&\;1&\;2&\;2\\\;2&\;2&\ldots&\;2&\;1&\;2\\-2&-2&\ldots&-2&-2&-1 \end{vmatrix}+(-1)^{n+1}\cdot(-2)\cdot\begin{vmatrix} \;1&\;1&\ldots&\;1&\;1&\;1 \\ 1&\;\;2&\ldots&\;2&\;2&\;2\\ \;2&1&...&\;2&\;2&\;2\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;2&\;2&\ldots&\;1&\;2&\;2\\\;2&\;2&\ldots&\;2&\;1&\;2\end{vmatrix}$$ $$=\begin{vmatrix} \;1&\;2&\ldots&\;2&\;2&\;2\\ \;2&1&\ldots&\;2&\;2&\;2\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;2&\;2&\ldots&\;1&\;2&\;2\\\;2&\;2&\ldots&\;2&\;1&\;2\\\;2&\;2&\ldots&\;2&\;2&\;1 \end{vmatrix}+2\cdot(-1)^{n}\cdot\begin{vmatrix} \;1&\;1&\ldots&\;1&\;1&\;1 \\ 1&\;\;2&\ldots&\;2&\;2&\;2\\ \;2&1&\ldots&\;2&\;2&\;2\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;2&\;2&\ldots&\;1&\;2&\;2\\\;2&\;2&\ldots&\;2&\;1&\;2\end{vmatrix}$$ I applied a formula derived earlier to the first sumand (where, instead of 1's on the main diagonal, there are parameters $a_k$ and x, whenever $i\ne j$ - under and above the diagonal): It looked like this: $$\color{blue}{\begin{vmatrix} \;a_1&\;x&\ldots&\;x&\;x&\;x\\ \;x&a_2&\ldots&\;x&\;x&\;x\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;x&\;x&\ldots&\;a_{n-2}&\;x&\;x\\\;x&\;x&\ldots&\;x&\;a_{n-1}&\;x\\\;x&\;x&\ldots&\;x&\;x&\;a_n\end{vmatrix} }$$ After subtracting the $\text{first row}$ from the rest of them: $$\color{blue}{\begin{vmatrix}\;a_1&\;x&\ldots&\;x&\;x&\;x\\ \;x-a_1&a_2-x&\ldots&\;0&\;0&\;0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;x-a_1&\;0&\ldots&\;a_{n-2}-x&\;0&\;0\\\;x-a_1&\;0&\ldots&\;0&\;a_{n-1}-x&\;0\\\;x-a_1&\;0&...&\;0&\;0&\;a_n-x\end{vmatrix}}$$ After knocking out the factor $a_j-x$ from every column: $$\color{blue}{\prod_{j=1}^{n} (a_j-x)\cdot\begin{vmatrix} \;\frac{a_1}{a_1-x}&\;\frac{x}{a_2-x}&\ldots&\;\frac{x}{a_{n-2}-x}&\;\frac{x}{a_{n-1}-x}&\;\frac{x}{a_n-x}\\ -1&1&\ldots&\;0&\;0&\;0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\-1&\;0&\ldots&\;1&\;0&\;0\\-1&\;0&\ldots&\;0&\;1&\;0\\-1&\;0&\ldots&\;0&\;0&\;1\end{vmatrix}}$$ After adding each column to the $\text{first}$ column we get the element: $$\frac{a_1}{a_1-x}+\sum_{j=2}^{n}\frac{x}{a_j-x}=\frac{a_1-x}{a_1-x} +\frac{x}{a_1-x}+x\sum_{j=2}^{n}\frac{1}{a_j-x}=1+x\sum_{j=1}^{n}\frac{1}{a_j-x}$$ on the position $1,1$ and $I_{n-1}$ inside the matrix. $$\color{blue}{\implies\det{X}=\prod_{j=1}^{n} (a_j-x)\;\cdot\;\left(1+x\sum_{j=1}^{n}\frac{1}{a_j-x}\right)}$$ In the task above, when I plugged $1,2$ and $(n-1)$ into the formula I got (for the first sumand): $$\prod_{k=1}^{n-1}(-1)\;\cdot\;(1-2(n-1))=(-1)^{n-1}(3-2n)$$ The second summand was the result of a transformation into $\text{lower triangular matrix}$ after subtracting each column from the next one: $$\begin{vmatrix} \;1&\;0&\ldots&\;0&\;0&\;0 \\ 1&\;\;1&\ldots&\;0&\;0&\;0\\ \;2&-1&\ldots&\;0&\;0&\;0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\;2&\;0&\ldots&-1&\;1&\;0\\\;2&\;0&\ldots&\;0&-1&1\end{vmatrix}$$ The $\text{product of the diagonal}$ is $1$. My final answer is (thanks to users in comments who noticed the arithmetic mistakes): $$\det A=(-1)^{n-1}(3-2n)+2\cdot(-1)^n=(2n-3)\cdot(-1)^n+2\cdot(-1)^n$$ $$\det A=(-1)^n(2n-3+2)=(-1)^n(2n-1)$$

Answer 1

Your answer isn't correct. Suppose $\mathbb F=\mathbb C$. Let $e_1=(1,0,\ldots,0,0)^T,\ v=\frac{1}{\sqrt{n-2}}(0,1,\ldots,1,0)^T$ and $e_n=(0,0,\ldots,0,1)^T$. Then $\{e_1,v,e_n\}$ is an orthonormal set of vectors and $$ A=-I-2\left(\sqrt{n-2}v+e_n\right)e_1^T+(e_1-2e_n)\left(\sqrt{n-2}\,v^T\right)+\left(e_1+\sqrt{n-2}v\right)e_n^T. $$ Therefore, if we extend $\{e_1,v,e_n\}$ to an orthonormal basis of $\mathbb C^n$, then $A$ is unitarily similar to $$ B=\pmatrix{-1&\sqrt{n-2}&1\\ -2\sqrt{n-2}&-1&\sqrt{n-2}\\ -2&-2\sqrt{n-2}&-1}\oplus (-I_{n-3}) $$ and hence $\det(A)=\det(B)=(1-2n)(-1)^{n-3}=(-1)^n(2n-1)$. By the method of universal identities, $\det(A)=(-1)^n(2n-1)$ over other fields as well.

We now return to your answer. In LaPlace expansion along the first column, the term containing $a_{n1}$ should be $\color{red}{(-1)^{n+1}}a_{n1}M_{n1}$. Therefore, the correct expansion should be \begin{aligned} \det(A) &=(-1)\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ -2&\cdots&\cdots&-2&-1 \end{vmatrix} +\color{red}{(-1)^{n+1}}(-2)\begin{vmatrix} 1&1&\cdots&1&1\\ 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2 \end{vmatrix}\\ &\phantom{}\\ &=\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ 2&\cdots&\cdots&2&1 \end{vmatrix} +2\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ 1&1&\cdots&1&1 \end{vmatrix}\\ &\phantom{}\\ &=\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ 2&\cdots&\cdots&2&1 \end{vmatrix} +\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ 2&2&\cdots&2&2 \end{vmatrix}\\ &\phantom{}\\ &=2\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ 2&\cdots&\cdots&2&1 \end{vmatrix} +\begin{vmatrix} 1&2&\cdots&2&2\\ 2&\ddots&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&2&\vdots\\ 2&\cdots&2&1&2\\ 0&0&\cdots&0&1 \end{vmatrix}\\ &=2\det(2E_{n-1}-I_{n-1})+\det(2E_{n-2}-I_{n-2})\\ &=2(-1)^{n-2}(2n-3)+(-1)^{n-3}(2n-5)\\ &=(-1)^n(2n-1), \end{aligned} where $E_k$ denotes the $k\times k$ all-one matrix such that $\det(2E_k-I_k)=(-1)^{k-1}(2k-1)$.

Answer 2

Here is a somewhat more elementary solution. Subtract the first row multiplied by $2$ from all other rows to obtain:

$$\begin{vmatrix} -1 &1&1&\cdots&1&1&1 \\ -2 &-1&0&\ldots&0&0&1\\ -2&0&-1&\cdots&\;0&\;0&\;1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ -2&0&0&\ldots&-1&0&1\\-2&0&0&\cdots&0&-1&1\\ -2&-2&-2&\cdots&-2&-2&-1 \end{vmatrix}_n = \begin{vmatrix} -1 &1&1&\cdots&1&1&1\\ 0 &-3&-2&\cdots&-2&-2&-1\\ 0 &-2&-3&\cdots&-2&-2&-1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0 &-2&-2&\ldots&-3&-2&-1\\ 0 &-2&-2&\ldots&-2&-3&-1\\ 0 &-4&-4&\ldots&-4&-4&-3 \end{vmatrix}_n$$

Now use Laplace expansion along the first column to obtain:

$$-\begin{vmatrix} -3&-2&-2&\cdots&-2&-2&-1\\ -2&-3&\;-2&\cdots&-2&-2&-1\\ -2&-2&\;-3&\cdots&-2&-2&-1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ -2&-2&-2&\ldots&-3&-2&-1\\ -2&-2&-2&\ldots&-2&-3&-1\\ -4&-4&-4&\ldots&-4&-4&-3 \end{vmatrix}_{n-1}$$

and then subtract the first row from rows $2, \ldots, n-2$ and subtract the first row multiplied by $3$ from the last.

$$-\begin{vmatrix} -3&-2&-2&\cdots&-2&-2&-1\\ 1&-1&0&\cdots&0&0&0\\ 1&0&-1&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 1&0&0&\ldots&-1&0&0\\ 1&0&0&\ldots&0&-1&0\\ 5&2&2&\ldots&2&2&0 \end{vmatrix}_{n-1}$$

Laplace expansion along the last column gives

$$(-1)^n\begin{vmatrix} 1&-1&0&\cdots&0&0&0\\ 1&0&-1&\cdots&0&0&0\\ 1&0&0&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 1&0&0&\ldots&0&-1&0\\ 1&0&0&\ldots&0&0&-1\\ 5&2&2&\ldots&2&2&2 \end{vmatrix}_{n-2}$$

Now add columns $2, \ldots, n-2$ to the first column. In the lower left corner we get $5+2(n-3) = 2n-1$.

$$(-1)^{n}\begin{vmatrix} 0&-1&0&\cdots&0&0&0\\ 0&0&-1&\cdots&0&0&0\\ 0&0&0&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\ldots&0&-1&0\\ 0&0&0&\ldots&0&0&-1\\ 2n-1&2&2&\ldots&2&2&2 \end{vmatrix}_{n-2}$$

Laplace expansion along the first column gives

$$(-1)^n(-1)^{n-1}(2n-1)\begin{vmatrix} -1&0&\cdots&0&0\\ 0&-1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\ldots&-1&0\\ 0&0&\ldots&0&-1\\ \end{vmatrix}_{n-3} = (-1)^n(-1)^{n-1}(-1)^{n-3}(2n-1)$$

which is equal to $(-1)^{3n-4}(2n-1)=(-1)^n(2n-1)$.