Question

Find the determinant of the (2n) x (2n)matrix$\mathit{A}\mathbf{=}\left[\begin{array}{cc}0& I_n\\ I_n& 0\end{array}\right]$

Short answer

Therefore, the determinant of (2n( x (2n) is given by,

$\det \mathrm{A}={\left(-1\right)}^{-\mathrm{n}}$.

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,

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

To find the determinant of the given matrix. 

We have a 2n x 2n block matrix

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

If we switch any two rows of A, the determinant will change its sign. First, we switch the first and (n+1) -th row, then$2^{nd}$and (n+2) -th row, etc. Lastly, we switch the (n-1) -th and (2n-1) --th row. Now, we've obtained the identity matrix$I_{2n}$. We had n row interchanges, therefore$1=\det {\mathrm{I}}_{2\mathrm{n}}={\left(-1\right)}^{\mathrm{n}}\det \mathrm{A}$.

So,

$$\begin{gathered} \det \mathrm{A}=\frac{1}{{\left(-1\right)}^{\mathrm{n}}} \\ \det \mathrm{A}={\left(-1\right)}^{\mathrm{n}} \end{gathered}$$