Overview:

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