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.
$\blacksquare$