Lecture 25: Invertibility of Matrices

Instructor: Nathan Johnston

Overview

A matrix ( A ) is invertible if there exists a matrix ( A^{-1} ) such that ( A A^{-1} = A^{-1} A = I ) (identity matrix).
Previously, checked both multiplication orders; today’s theorem simplifies this.

Statement: For a square matrix ( A ), if there exists a matrix ( B ) such that either ( AB = I ) or ( BA = I ), then ( B ) is the inverse of ( A ).
Implication: You only need to check one side for the identity matrix to confirm invertibility.
Proof Outline:
- Assume ( BA = I ).
- Show that the system ( A\mathbf{x} = 0 ) has a unique solution, implying ( A ) is invertible.
- Multiply ( BA = I ) by ( A^{-1} ) to confirm ( B = A^{-1} ).
- Same logic applies if ( AB = I ).

Explicit Formula: For a 2x2 matrix ( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} ), it is invertible if ( ad - bc \neq 0 ).
Inverse Formula: If invertible, ( A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} ).
Larger matrices have a formula but are impractical; use row-reduction instead.

Verification:
- Multiply ( A ) by ( A^{-1} ) using the formula and confirm identity matrix.
Case Analysis:
- Case 1: Zero row or column implies not invertible.
- Case 2: Proportional rows imply not invertible.
- If ( ad - bc = 0 ), matrix is not invertible.

Matrix: ( \begin{pmatrix} 3 & -2 \ 1 & 4 \end{pmatrix} )
- ( ad-bc = 14 \neq 0 ), matrix is invertible.
- Inverse: ( \frac{1}{14} \begin{pmatrix} 4 & 2 \ -1 & 3 \end{pmatrix} ).
Matrix: ( \begin{pmatrix} 2 & -3 \ -4 & 6 \end{pmatrix} )
- ( ad-bc = 0 ), matrix is not invertible.
Matrix: ( \begin{pmatrix} \pi & 3.5 \ -2 & 6 \end{pmatrix} )
- ( ad-bc = 6\pi + 7 \neq 0 ), matrix is invertible.
- Inverse: ( \frac{1}{6\pi + 7} \begin{pmatrix} 6 & -3.5 \ 2 & \pi \end{pmatrix} ).