Matrices 6: Inverses of 2 × 2 matrices

Overview:
  1. Introduction to matrices
  2. Adding and subtracting matrices
  3. Multiplying matrices
  4. 2 × 2 Matrices and linear transformations
  5. Determinants of 2 × 2 matrices
  6. Inverses of 2 × 2 matrices
  7. Invariant points and lines in 2 dimensions
  8. 3 × 3 Matrices and linear transformations
  9. Determinants of 3 × 3 matrices
  10. Inverses of 3 × 3 matrices
  11. Matrices and simultaneous equations

 Part 6: Inverses of 2 × 2 matrices


Given two matrices A and B, if AB = I, the identity matrix, then B is the inverse of A. We can denote the inverse of A as A-1 i.e. B = A-1.

A square matrix M has an inverse, denoted M-1, if and only if its |M| ≠ 0. If the determinant of M is 0, then M has no inverse.

Given a matrix M and its inverse, M-1, the following will be true:

  • MM-1 = I
  • M-1M = I

The inverse of a 2 × 2 matrix \(\textsf{M}\) = \(\begin{pmatrix} a & b\\c & d\end{pmatrix}\) can be found (where it exists) as follows:

\(\textsf{M}^{-1}\) = \(\dfrac{1}{\begin{vmatrix} \textsf{M} \end{vmatrix}} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}\) = \(\dfrac{1}{ad-bc} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}\).

Extension 1: Verify that MM-1 = I.

Extension 2: Verify that M-1M = I.

Extension 3: Find the determinant of M-1.

Questions
  1. Can you see why (at least for a 2 \(\times\) 2 matrix), that an inverse can’t exist if the matrix has determinant 0?
  2. Show that \(|\textsf{M}^{-1}| = \dfrac{1}{\textsf{|M|}}\).
Answers
  1. \(\textsf{M}^{-1} = \dfrac{1}{\begin{vmatrix} \textsf{M} \end{vmatrix}} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}\), so if \(\textsf{|M|} = 0\), the inverse would be \(\dfrac{1}{0} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}\), which is a problem because \(\dfrac{1}{0}\) is not defined.

     

  2. If \(\textsf{M}\) = \(\begin{pmatrix} a & b\\c & d\end{pmatrix}\), then \(\textsf{M}^{-1}\) = \(\dfrac{1}{\begin{vmatrix} \textsf{M} \end{vmatrix}} \begin{pmatrix} d & -b\\-c & a\end{pmatrix} = \begin{pmatrix} \dfrac{d}{\textsf{|M|}} & \dfrac{-b}{\textsf{|M|}}\\\dfrac{-c}{\textsf{|M|}} & \dfrac{a}{\textsf{|M|}}\end{pmatrix}\).

    So \(|\textsf{M}^{-1}| = \dfrac{ad}{\textsf{|M|}^{2}} – \dfrac{(-b)(-c)}{\textsf{|M|}^{2}}=\dfrac{ad-bc}{\textsf{|M|}^{2}}=\dfrac{\textsf{|M|}}{\textsf{|M|}^{2}}=\dfrac{1}{\textsf{|M|}}\).