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 unit matrix of order $n$ over $R$ is equal to $1_R$.

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

We have that:


 * $\map \det {\mathbf I_2} = \begin {vmatrix}

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

Suppose $\map \det {\mathbf I_n} = 1_R$.

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:
 * $\map \det {\mathbf I_{n + 1} } = \map \det {\mathbf I_n}$

Hence the result, by induction.