A 1 by 1 matrix ` A` = (

Assume the 2 by 2 matrix

= [A |
a |
b |
] |

c |
d |

has

= [B |
x |
z |
] |

y |
w |

as the inverse. Then from ** AB** =

`ax` + `by` = 1, `cx` + `dy` = 0, `az` + `bw` = 0, `cz` + `dw` = 1.

The first two is a system of linear equations with ` A` as the coefficient matrix and

(`ad` - `bc`)`x` = `d`, (`ad` - `bc`)`y` = - `c`.

Similarly, from the 3rd and the 4th equations we get

(`ad` - `bc`)`z` = - `b`, (`ad` - `bc`)`w` = `a`.

Thus we conclude the ⇐ direction of the following result.

A 2 by 2 matrix [ | a |
b |
] is invertible ⇔ ad - bc ≠ 0. Moreover, the inverse is |

c |
d |

1 | [ | d |
-b |
] |

ad - bc |
-c |
a |

For the ⇒ direction, see this exercise.

Example We have

[ | 1 | 3 | ]^{-1} =
(1×4 - 3×2)^{-1}[ |
4 | -3 | ]^{-1} = [ |
-2 | 3/2 | ] |

2 | 4 | -2 | 1 | 1 | -1/2 |

We verify our computation as follows.

[ | 1 | 3 | ] [ | -2 | 3/2 | ] = [ | 1×(-2) + 3×1 | 1×3/2 + 3×(-1/2) | ] = [ | 1 | 0 | ] |

2 | 4 | 1 | -1/2 | 2×(-2) + 4×1 | 2×3/2 + 4×(-1/2) | 0 | 1 |

[ | -2 | 3/2 | ] [ | 1 | 3 | ] = [ | (-2)×1 + (3/2)×2 | (-2)×3 + (3/2)×4 | ] = [ | 1 | 0 | ] |

1 | -1/2 | 2 | 4 | 1×1 + (-1/2)×2 | 1×3 + (-1/2)×4 | 0 | 1 |

The inverse also tells us that the system

x_{1} |
+ 3x_{2} |
= | 1 |

2x_{1} |
+ 4x_{2} |
= | -4 |

has

[ | x_{1} |
] = [ | 1 | 3 | ]^{-1} [ |
1 | ] = [ | -2 | 3/2 | ] [ | 1 | ] = [ | 4 | ] |

x_{2} |
2 | 4 | -4 | 1 | -1/2 | 4 | -1 |

as the unique solution.

Example The following matrices

[ | 2 | 4 | ] |

-1 | -2 |

[ | 1 | 5 | ] |

3 | 15 |

are not invertible because 2×(-2) - 4×(-1) = 0, 1×15 - 5×3 = 0.

The formula for the inverse of 2 by 2 matrices can be generalized to bigger matrices. However, the amount of computation involved in such a formula will become too big to be practical.