Proof
- The result of the Multiplication of Square Matrix A and the inverse A-1 and also the inverse A-1 and Square Matrix A will give us the Unit Matrix
\[\Large
A \cdot A^{-1} =
\begin{bmatrix}
\textcolor{#FF0000}{1} & 0 & 0 \\
0 & \textcolor{#FF0000}{1} & 0 \\
0 & 0 & \textcolor{#FF0000}{1} \\
\end{bmatrix}
\]
\[\Large
A^{-1} \cdot A =
\begin{bmatrix}
\textcolor{#FF0000}{1} & 0 & 0 \\
0 & \textcolor{#FF0000}{1} & 0 \\
0 & 0 & \textcolor{#FF0000}{1} \\
\end{bmatrix}
\]
- This demonstrates that the product of a square matrix and its inverse, regardless of the order of multiplication, is the unit matrix.
- Dont know what is a Unit Matrix? check out Types of Matrices Lesson
\[\Large
\therefore \, \color{blue}{A} \cdot \color{red}{A^{-1}} = \color{green}{I}
\]
\[\Large
\&
\]
\[\Large
\color{red}{A^{-1}} \cdot \color{blue}{A} = \color{green}{I}
\]
If \( A \) is a square matrix and \( A^{-1} \) is its inverse, what is the result of \( A \cdot A^{-1} \) or \( A^{-1} \cdot A \)?