Determinant of Unit Matrix

Theorem
Let $$R$$ be a ring with unity whose zero is $$0_R$$ and whose unity is $$1_R$$.

The determinant of the identity matrix of order $$n$$ over $$R$$ is equal to $$1_R$$.

Proof
Let $$\mathbf I_n$$ denote the identity matrix of order $$n$$ over $$R$$

We have that:


 * $$\det \left({\mathbf I_2}\right) = \begin{vmatrix}

1_R & 0_R \\ 0_R & 1_R \end{vmatrix} = 1_R \cdot 1_R - 0_R \cdot 0_R = 1_R$$

Now suppose $$\det \left({\mathbf I_n}\right) = 1_R$$.

We have that $$\mathbf I_{n+1} = \begin{bmatrix} 1_R & 0_R \\ 0_R & \mathbf I_n \end{bmatrix}$$

Hence from Determinant with Unit Element in Otherwise Zero Row we have that $$\det \left({\mathbf I_{n+1}}\right) = \det \left({\mathbf I_n}\right)$$.

Hence the result, by induction.