Question

Let ${\mathit{P}}_{\mathbf{n}}$ be the $\mathit{n}\mathbf{\times }\mathit{n}$ matrix whose entries are all ones, except for zeros directly below the main diagonal; for example,

role="math" localid="1659508976827" ${\mathit{P}}_{\mathbf{5}}\mathbf{\left[}\begin{array}{ccccc}\mathbf{1}& \mathbf{1}& \mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{0}& \mathbf{1}& \mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{0}& \mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{1}& \mathbf{0}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{1}& \mathbf{1}& \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right]}$

Find the determinant of ${\mathit{P}}_{\mathbf{n}}$ .

Short answer

Therefore, the determinant of the given matrix is given by,

$det{P}_n=1,\forall n\in \square .$

Step-by-step solution

Definition. 

A determinant is a unique number associatedwith a square matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given. 

Given matrix,

$P_5\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 1& 1& 1\\ 1& 0& 1& 1& 1\\ 1& 1& 0& 1& 1\\ 1& 1& 1& 0& 1\end{array}\right]$

To find determinant.

Using the Laplace expansion along the$n^{th}$row, we see that the first$n-1$addends are 0, because the corresponding minors for the first 2 elements are 0, since each of them has $n-2$ columns with all entries 1; and the entry$n^{th}$ in row and $\left(n-1\right)$ column is 0.

Thus, we're only left with $1.det{P}_{n-1}$. Since$det{P}_n=1,\forall n\in \square$ , we have .