Question

Find the determinant of the$\left(2n\right)\mathbf{\times }\left(2n\right)$matrix

$\mathit{A}\mathbf{=}\left[\begin{array}{cc}0& l_n\\ I_n& 0\end{array}\right]$.

Short answer

Therefore, the determinant of the given matrix is given by, $detA={\left(-1\right)}^n$.

Step-by-step solution

Definition

A determinant is a unique number associated witha 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,

$A=\left[\begin{array}{cc}0& I_n\\ I_n& 0\end{array}\right]$

To find the determinant

To obtain the identity matrix $I_{2n}$, we first switch the first and $\left(n+1\right)$-th row, then we switch the second and $\left(n+2\right)$-th row, etc.

Lastly, we switch the n-th and 2n -th row.

We had n row interchanges, so $detA={\left(-1\right)}^n$.