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Let $$A = \begin{pmatrix} 1 & 1 & 0 & 0 &0 &... & 0 \\ -1 & 1 & 0 & 0 &0&...&0 \\ -1 & 0 & 1 & 0 &0&...&0 \\ -1 & 0 & 0 & 1 &0&...&0 \\ -1 & 0 & 0 & 0 &1&...&0 \\ \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&\vdots \\ -1 & 0 & 0 & 0 &0&...&1\end{pmatrix}$$

be an $n \times n$ matrix consisting of $1$'s on its diagonal, a $1$ in the entry located on row $1$ and column $2$, and $-1$'s from the second entry of the first column all the way to $n$. Prove that the determinant of $A$ is always $2$.

How should I begin this? I tried taking the determinants of the matrix when $n=2, 3, $ and $ 4$ and saw that they were all $2$, but am not sure how to proceed.

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    $\begingroup$ Hint: Try performing a Laplace expansion on the first row. $\endgroup$
    – Elliot Yu
    Commented Jul 19, 2022 at 2:06
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    $\begingroup$ Or perhaps try an inductive proof by expanding on the last column. $\endgroup$
    – Elliot Yu
    Commented Jul 19, 2022 at 2:07
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    $\begingroup$ Subtract the second row from the first one, you get a lower triangular matrix whose main diagonal is $(2,1,1,\ldots)$. $\endgroup$
    – user1551
    Commented Jul 19, 2022 at 10:43
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    $\begingroup$ note your matrix is block lower triangular of the form $A = \begin{pmatrix} B & \mathbf 0 \\ * & I \end{pmatrix}$ where $B = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ hence $det\big(A\big) = \det\big(B\big)\det\big(I\big) = 2 \cdot 1 = 2$ $\endgroup$ Commented Jul 20, 2022 at 4:28

1 Answer 1

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Determinants are unchanged if you add a row, or a multiple thereof, to another.

Hence, add the first row to every other row.

$$\begin{pmatrix} 1 & 1 & 0 & 0 &0 & \cdots & 0 \\\ 0 & 2 & 0 & 0 &0& \cdots&0 \\\ 0 & 1 & 1 & 0 &0& \cdots&0 \\\ 0 & 1 & 0 & 1 &0& \cdots&0 \\\ 0 & 1 & 0 & 0 &1& \cdots&0 \\\ \vdots & \vdots & \vdots& \vdots &\vdots & \ddots &\vdots \\\ 0 & 1 & 0 & 0 &0& \cdots&1\end{pmatrix}$$

Now add $-1/2$ times the second row to the first row:

$$\begin{pmatrix} 1 & 0 & 0 & 0 &0 & \cdots & 0 \\\ 0 & 2 & 0 & 0 &0& \cdots&0 \\\ 0 & 1 & 1 & 0 &0& \cdots&0 \\\ 0 & 1 & 0 & 1 &0& \cdots&0 \\\ 0 & 1 & 0 & 0 &1& \cdots&0 \\\ \vdots & \vdots & \vdots& \vdots &\vdots & \ddots &\vdots \\\ 0 & 1 & 0 & 0 &0& \cdots &1\end{pmatrix}$$

This is a triangular matrix; the determinant of such a matrix is the product of its diagonal entries. Hence, the determinant is $2$.


Addendum:

As noted in the comments, a simpler solution in the same spirit is to just subtract the second row from the first. This gives

$$\begin{pmatrix} 2 & 0 & 0 & 0 &0 &... & 0 \\ -1 & 1 & 0 & 0 &0&...&0 \\ -1 & 0 & 1 & 0 &0&...&0 \\ -1 & 0 & 0 & 1 &0&...&0 \\ -1 & 0 & 0 & 0 &1&...&0 \\ \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&\vdots \\ -1 & 0 & 0 & 0 &0&...&1\end{pmatrix}$$

This too is lower-triangular, so the determinant is the product of the diagonal entries, which is clearly $2$.

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    $\begingroup$ Isn't it enough to subtract the second row from the first in the original matrix? $\endgroup$ Commented Jul 19, 2022 at 10:58
  • $\begingroup$ Oh, yeah, that would be a bit simpler yeah $\endgroup$ Commented Jul 19, 2022 at 19:41

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