An invertible matrix *A* is called a row equivalent to an identity matrix, and we can this matrix by understanding the row reduction of *A* to *I*.

A n x n matrix *A* is invertible if and only if *A* is row equivalent to *I _{n}*, and in this case, any sequence of elementary row operations that reduces

*A*to

*I*also transforms

_{n}*I*into A

_{n}^{-1}.

Proof: Suppose that *A* is an invertible matrix.Then, since the equation Ax =b has a solution for each b, *A* has a pivot position in every row.Because *A* is square, the *n* pivot positions must be on the diagonal, which implies that the reduced echelon form of *A* is *I _{n}. A˜I_{n}*.Then, since each step of the row reduction of

*A*corresponds to left-multiplication by an elementary matrix, there exist elementary matrices

*E*

_{1}, …,

*E*such that A~E

_{p}_{1}A~E

_{2}(E

_{1}A)~…~E

_{p}(E

_{p-1}…E

_{1}A)=I

_{n}

Therefore, E_{p}…E_{1}A = I_{n}. Since the product *E _{p}*…

*E*

_{1}of invertible matrices is invertible, the equation leads to (E

_{p}…E

_{1})

^{-1}(E

_{p}…E

_{1})A = (E

_{p}…E

_{1})

^{-1}I

_{n}

A = (E_{p}…E_{1})^{-1}

When a matrix *A* is invertible, the equation A^{-1} Ax = x can be applied as statement for linear transformations. See the following figure:

The inverse of a matrix is defined by **AB** = I = **BA** if and only if **A** is the ** inverse** of

**B**.

We then write: **AA**^{-1} = **A**^{-1}**A **= 1 = **BB**^{-1}=** B**^{-1}**B. **Consider the general matrix expression below:

**AX** = **B**

**A**^{-1}**AX** = **A**^{-1}**B**

**A**^{-1}**AX** = **A**^{-1}**B**

1 **X** = **A**^{-1}**B**

**X** = **A**^{-1}**B**

If A and B are invertibile matrices having same order, then (AB)^{-1 }= B^{-1} A^{-1}.

From the definition of inverse matrix, we know that:

(AB) (AB)^{–1} =1

A^{–1} (AB) (AB)^{–1} =A^{–1}I (Pre multiplying both sides by A^{–1})

(A_{–1}A) B (AB)_{–1} =A_{–1} (Since A_{–1} I = A_{–1})

IB (AB)^{–1} =A^{–1} or B (AB)^{–1} =A^{–1}

B^{–1} B (AB)^{–1} =B^{–1} A^{–1}

I (AB)^{–1} =B–^{1} A^{–1}

Thus, (AB)^{–1} =B^{–1} A^{–1}

**Uniqueness of inverse: If an inverse of a square matrix exists, it is unique.**

Proof: Let A = [a_{ij}] to be a square matrix havingan order of m. Let B and C be two possible inverses of A. If the inverse is unique, then B = C

Therefore, AB = BA = I and AC = CA = I since both B and C are inverse of A.

Hence,

B = BI = B(AC) =(BA)C = IC = C