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Non-Singular Matrices and Unique Inverses

An invertible square matrix is also called non-singular. A non-invertible square matrix is singular. The inverse $A^{-1}$ satisfies $A^{-1}A = AA^{-1} = I$ .

The inverse is unique: if $L$ is a left inverse and $R$ is a right inverse of $A$ , then $L = R$ (and so the two-sided inverse is unique). Proof: $L = LI = L(AR) = (LA)R = IR = R$ .

Key inverse properties: $(A^{-1})^{-1} = A$ , $(AB)^{-1} = B^{-1}A^{-1}$ (note order reversal), $(A^T)^{-1} = (A^{-1})^T$ .

Formal View

Theorem 4.11 — Uniqueness and Properties of Inverse

A \in \mathbb{R}^{n \times n}

is invertible, its inverse

A^{-1}

is unique and satisfies: \begin{enumerate} \item

(A^{-1})^{-1} = A

\item

(AB)^{-1} = B^{-1}A^{-1}

for invertible

B

\item

(A^T)^{-1} = (A^{-1})^T

\end{enumerate}

Why This Matters

Knowing inverse properties lets us manipulate matrix equations like scalar equations.

Solving $AX = B$ : $X = A^{-1}B$ (but computing $A^{-1}$ explicitly is expensive — use LU instead)
$(AB)^{-1} = B^{-1}A^{-1}$ is the "reverse-order law," critical in signal processing and control
The formula $(A^T A)^{-1} A^T$ (Moore-Penrose pseudoinverse) appears in least squares

Learning Resources

Inverse of a matrix

Khan Academy

Computing and using matrix inverses.

11 min

Quiz

Question 1

If $A$ and $B$ are invertible, $(AB)^{-1}$ equals:

Common Mistakes

Writing $(AB)^{-1} = A^{-1}B^{-1}$ — the correct order reverses: $B^{-1}A^{-1}$ .
Computing $A^{-1}$ explicitly when solving $A\mathbf{x} = \mathbf{b}$ — it's better to use LU decomposition.