Inverse of a Matrix

Before we look at what an inverse of a matrix is, please read about Identity Matrices, that is a square matrix who diagonal entries are all 1's and all other entries are 0's.

# Inverse of a Matrix

 Definition: If $A$ and $B$ are both square $n \times n$ matrices such that $AB = BA = I_n$, then $B$ is the Inverse Matrix of $A$ commonly denoted $B = A^{-1}$. If no such matrix $B$ exists such that $AB = BA = I_n$, then $A$ is said to be Singular or Noninvertible.

For example, consider the following matrices $A$ and $B$:

(1)
\begin{align} A = \begin{bmatrix} 2& 3\\ -1 & -3 \end{bmatrix} \quad , \quad B = \begin{bmatrix} 1 & 1\\ -\frac{1}{3} & -\frac{2}{3} \end{bmatrix} \end{align}

If we calculate $AB$, we end up getting the identity matrix, that is:

(2)
\begin{align} AB = \begin{bmatrix} 2& 3\\ -1 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1\\ -\frac{1}{3} & -\frac{2}{3} \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = I \end{align}

Therefore we can say that matrix $B$ is the inverse of $A$, that is $B = A^{-1}$. Later, we will find methods to find the inverse of a matrix (if it exists) and look at their practical applications.

# Properties of the Inverse of a Matrix

• a. If $A$ is invertible and $B$ and $C$ are inverses of $A$, then $B = C$.
• b. If $A$ is invertible, then $A^{-1}$ is also invertible and $(A^{-1})^{-1} = A$.
• c. If $A$ and $B$ are invertible matrices of size $n \times n$, then their product $AB$ is invertible and $(AB)^{-1} = B^{-1}A^{-1}$.
• d. If $A$ is invertible, then the tranpose of $A$ is also invertible and $(A^T)^{-1} = (A^{-1})^T$.

## Proof of the Properties of the Inverse of a Matrix

• Proof of (a): Since $B$ is the inverse of $A$, it follows that $AB = I$. We will now multiply both sides of this equation on the left by $C$, that is $C(AB) = CI = C$. Recall that $C(AB) = (CA)B$, and therefore it follows that $(CA)B = C$. Since $C$ is also an inverse of $A$, it follows that $B = C$. $\blacksquare$
• Proof of (b): We know that $AA^{-1} = A^{-1}A = I$. Therefore it follows that $(A^{-1})^{-1} = A$. $\blacksquare$
• Proof of (c): First assume that $(AB)^{-1} = B^{-1}A^{-1}$. We will now multiply both sides on the left by $AB$:
(3)
\begin{align} (AB)^{-1} = B^{-1}A^{-1} \\ (AB) (AB)^{-1} = (AB)B^{-1}A^{-1} \\ I = A(BB^{-1})A^{-1} \\ I = AIA^{-1} \\ I = AA^{-1} \\ I = I \end{align}
• Therefore our original assumption that $(AB)^{-1} = B^{-1}A^{-1}$ was true. $\blacksquare$