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.