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Q29E

Expert-verifiedFound in: Page 289

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**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 ${{\mathit{P}}}_{{\mathbf{n}}}$** **.**

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

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

A determinant is a unique number associated with 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 matrix,

${{P}}_{{5}}\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 1& 1& 1\\ 1& 0& 1& 1& 1\\ 1& 1& 0& 1& 1\\ 1& 1& 1& 0& 1\end{array}\right]$

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 .

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