Mathematics · Ch 3 — Matrices

Invertible Matrices

3.7

Invertible Matrices

3.7 Invertible Matrices

The Concept of Inverse

Every non-zero number has a reciprocal, whose product with the number is 1. For matrices there is an analogous idea: the inverse of a square matrix, whose product with the matrix is the identity matrix $I$ (the matrix equivalent of 1).

Important

Only square matrices can have inverses. For a rectangular matrix, the products $AB$ and $BA$ cannot both be defined and equal.

Definition of Inverse

Let $A$ be a square matrix of order $m$ . If there exists a square matrix $B$ of the same order such that

AB = BA = I

then $B$ is the inverse of $A$ , written $B = A^{-1}$ , and $A$ is invertible.

Note

$A^{-1}$ is read "A inverse". It does not mean $\frac{1}{A}$ — division is not defined for matrices.

Example

Let $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$ .

AB = \begin{bmatrix} 4-3 & -6+6 \\ 2-2 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

BA = \begin{bmatrix} 4-3 & 6-6 \\ -2+2 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

Since $AB = BA = I$ , we have $B = A^{-1}$ and also $A = B^{-1}$ .

Tip

If $B$ is the inverse of $A$ , then $A$ is automatically the inverse of $B$ — the relationship is mutual.

Theorem 3: Uniqueness of Inverse

Statement: The inverse of a square matrix, if it exists, is unique.

Proof: Suppose $A$ has two inverses $B$ and $C$ , so $AB = BA = I$ and $AC = CA = I$ . Then

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

using associativity and $BA = I$ . Hence $B = C$ : the inverse is unique, so we may speak of the inverse of $A$ .

Theorem 4: Inverse of a Product

Statement: If $A$ and $B$ are invertible matrices of the same order, then

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

Proof: Start from $(AB)(AB)^{-1} = I$ and pre-multiply by $A^{-1}$ :

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

Now pre-multiply by $B^{-1}$ :

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

Watch out

A common mistake is $(AB)^{-1} = A^{-1}B^{-1}$ . The order reverses — like removing shoes before socks. This extends to more factors: $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ .

›Proof

Verification. $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = I$ , and $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}B = I$ . Both products equal $I$ , so $B^{-1}A^{-1}$ satisfies the definition of $(AB)^{-1}$ .

Check Your Understanding …

Definition 6Invertible Matrices

Definition of Invertible Matrices

Let $A$ be a square matrix of order $m$ (i.e., it has $m$ rows and $m$ columns).

If there exists another square matrix $B$ of the same order $m$ such that

AB = BA = I

(where $I$ is the identity matrix of order $m$ ), then:

$B$ is called the inverse of $A$ , written as $B = A^{-1}$ .
$A$ is said to be invertible (or non-singular).

Key conditions from the textbook:

Both $A$ and $B$ must be square matrices of the same order — a rectangular matrix cannot have an inverse.
The products $AB$ and $BA$ must both equal the identity matrix $I$ .
If $B$ is the inverse of $A$ , then $A$ is also the inverse of $B$ (i.e., $A = B^{-1}$ ).
The inverse of a square matrix, if it exists, is unique (Theorem 3).

Intuition

Think of a number: the inverse of $2$ is $\frac{1}{2}$ because $2 \times \frac{1}{2} = 1$ .

For matrices, the identity matrix $I$ plays the role of "1". So $A^{-1}$ is the matrix that "undoes" $A$ — multiplying them in either order gives back $I$ .

Tiny Concrete Example

Let

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

Then …

Theorem 3

Theorem 4 (Reversal Law for Inverses)

If $A$ and $B$ are invertible matrices of the same order, then the product $AB$ is also invertible, and its inverse is given by the reverse-order product of the individual inverses:

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

Hypotheses:

$A$ and $B$ are square matrices of the same order (say $n \times n$ ).
Both $A$ and $B$ are invertible — that is, there exist matrices $A^{-1}$ and $B^{-1}$ such that $AA^{-1} = A^{-1}A = I$ and $BB^{-1} = B^{-1}B = I$ , where $I$ is the identity matrix of order $n$ .

Important

The order matters: the inverse of a product is the product of the inverses in reverse order. This is not a commutative law — you cannot swap $A$ and $B$ unless they happen to commute, which is rare.

Proof

›Proof

We start from the definition of the inverse. To show that $B^{-1}A^{-1}$ is the inverse of $AB$ , we must verify that

$(AB)(B^{-1}A^{-1}) = I \quad \text{and} \quad (B^{-1}A^{-1})(AB) = I.$

Step 1 — Multiply $AB$ on the right by $B^{-1}A^{-1}$ :

$(AB)(B^{-1}A^{-1}) = A\,(B B^{-1})\,A^{-1}$

Here we used the associative property of matrix multiplication: $(AB)(B^{-1}A^{-1}) = A(B(B^{-1}A^{-1})) = A((BB^{-1})A^{-1})$ .

Since $B B^{-1} = I$ , this becomes

$A\,I\,A^{-1} = A A^{-1} = I.$

Step 2 — Multiply $AB$ on the left by $B^{-1}A^{-1}$ :

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

Again by associativity: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}(AB)) = B^{-1}((A^{-1}A)B)$ .

Since $A^{-1}A = I$ , we get

$B^{-1} I B = B^{-1}B = I.$

Both products equal the identity matrix $I$ . Therefore, by definition, $B^{-1}A^{-1}$ is the inverse of $AB$ , and we write

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

Why This Matters …

Theorem 4

Theorem 4 (Reversal Law for Inverses)

If $A$ and $B$ are invertible matrices of the same order, then the product $AB$ is also invertible, and its inverse is given by the reverse-order product of the individual inverses:

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

Hypotheses:

$A$ and $B$ are square matrices of the same order (say $n \times n$ ).
Both $A$ and $B$ are invertible — that is, there exist matrices $A^{-1}$ and $B^{-1}$ such that $AA^{-1} = A^{-1}A = I$ and $BB^{-1} = B^{-1}B = I$ , where $I$ is the identity matrix of order $n$ .

Important

Proof

›Proof

We start from the definition of the inverse. To show that $B^{-1}A^{-1}$ is the inverse of $AB$ , we must verify that

$(AB)(B^{-1}A^{-1}) = I \quad \text{and} \quad (B^{-1}A^{-1})(AB) = I.$

Step 1 — Multiply $AB$ on the right by $B^{-1}A^{-1}$ :

$(AB)(B^{-1}A^{-1}) = A\,(B B^{-1})\,A^{-1}$

Here we used the associative property of matrix multiplication: $(AB)(B^{-1}A^{-1}) = A(B(B^{-1}A^{-1})) = A((BB^{-1})A^{-1})$ .

Since $B B^{-1} = I$ , this becomes

$A\,I\,A^{-1} = A A^{-1} = I.$

Step 2 — Multiply $AB$ on the left by $B^{-1}A^{-1}$ :

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

Again by associativity: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}(AB)) = B^{-1}((A^{-1}A)B)$ .

Since $A^{-1}A = I$ , we get

$B^{-1} I B = B^{-1}B = I.$

Both products equal the identity matrix $I$ . Therefore, by definition, $B^{-1}A^{-1}$ is the inverse of $AB$ , and we write

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

Why This Matters …

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