# Why $\det\begin{pmatrix} 1&1&0&...&0 \\ -1 & 1 & 0 &...&0 \\ -1 & 0 & 1 &...&0 \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ -1&0&0&...&1\end{pmatrix}=2$?

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.

• Hint: Try performing a Laplace expansion on the first row. Commented Jul 19, 2022 at 2:06
• Or perhaps try an inductive proof by expanding on the last column. Commented Jul 19, 2022 at 2:07
• Subtract the second row from the first one, you get a lower triangular matrix whose main diagonal is $(2,1,1,\ldots)$. Commented Jul 19, 2022 at 10:43
• 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 $det\big(A\big) = \det\big(B\big)\det\big(I\big) = 2 \cdot 1 = 2$ Commented Jul 20, 2022 at 4:28

## 1 Answer

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 $$2$$.

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