Unit Matrix is its own Inverse

Theorem
The inverse of the unit matrix $\mathbf I_n$ of order $n$ is $\mathbf I_n$.

That is, a unit matrix it its own inverse.

Proof
By definition, a unit matrix is a diagonal matrix.

From Inverse of Diagonal Matrix, the inverse of a diagonal matrix:


 * $\mathbf D = \begin{bmatrix}

a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$

is the diagonal matrix:


 * $\mathbf D^{-1} = \begin{bmatrix}

\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\ 0 & \dfrac 1 {a_{22}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\ \end{bmatrix}$

When $\mathbf D$ is the unit matrix $\mathbf I_n$, all elements $a_{kk}$ are equal to $1$:


 * $\mathbf I_n = \begin{bmatrix}

1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}$

Hence:


 * $\mathbf I_n^{-1} = \begin{bmatrix}

\dfrac 1 1 & 0 & \cdots & 0 \\ 0 & \dfrac 1 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \dfrac 1 1 \\ \end{bmatrix}$

Hence the result.