Inverse of a Matrix

DEFINITION 1.2.13 (Inverse of a Matrix)   Let $A$ be a square matrix of order $n.$

1. A square matrix $B$ is said to be a LEFT INVERSE of $A$ if $B A = I_n.$
2. A square matrix $C$ is called a RIGHT INVERSE of $A,$ if $A C = I_n.$
3. A matrix $A$ is said to be INVERTIBLE (or is said to have an INVERSE) if there exists a matrix $B$ such that $A B = B A = I_n.$

LEMMA 1.2.14   Let $A$ be an $n \times n$ matrix. Suppose that there exist $n \times n$ matrices $B$ and $C$ such that $A B = I_n$ and $C A = I_n,$ then $B = C.$

Proof. Note that

$$C = C I_n = C( A B) = (C A) B = I_n B = B.$$

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Remark 1.2.15  

1. From the above lemma, we observe that if a matrix $A$ is invertible, then the inverse is unique.
2. As the inverse of a matrix $A$ is unique, we denote it by $A^{-1}.$ That is, $A A^{-1} = A^{-1} A = I.$

THEOREM 1.2.16   Let $A$ and $B$ be two matrices with inverses $A^{-1}$ and $B^{-1},$ respectively. Then

1. $(A^{-1})^{-1}= A.$
2. $( A B )^{-1} = B^{-1} A^{-1}.$
3. $(A^t)^{-1} = (A^{-1})^t.$

Proof. Proof of Part 1.
By definition $A A^{-1} = A^{-1} A = I.$ Hence, if we denote $A^{-1}$ by $B,$ then we get $A B = B A = I.$ Thus, the definition, implies $B^{-1} = A,$ or equivalently $(A^{-1})^{-1}= A.$

Proof of Part 2.
Verify that $(A B) (B^{-1} A^{-1}) = I = (B^{-1} A^{-1}) (A B).$

Proof of Part 3.
We know $A A^{-1} = A^{-1} A = I.$ Taking transpose, we get

$$(A A^{-1})^t = (A^{-1} A)^t = I^t \Longleftrightarrow (A^{-1})^t A^t = A^t (A^{-1})^t = I.$$

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EXERCISE 1.2.17  

1. Let $A_1, A_2, \ldots, A_r$ be invertible matrices. Prove that the product $A_1 A_2 \cdots A_r$ is also an invertible matrix.
2. Let $A$ be an inveritble matrix. Then prove that $A$ cannot have a row or column consisting of only zeros.
3. Let $A=[a_{ij}]$ be an invertible matrix and let $p$ be a nonzero real number. Then determine the inverse of the matrix $B = [p^{i-j} a_{ij}]$ .