## Inverse matrix

Let $$A, M, N \in \mathbb{F}^{n\times n}$$ where $$\mathbb{F}$$ denotes a field. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$.

The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Hence, it could very well be that $$AB = I_n$$ but $$BA$$ is something else.

Interestingly, it turns out that left inverses are also right inverses and vice versa. You can see a proof of this here. What follows is a proof of the following easier result:

If $$MA = I$$ and $$AN = I$$, then $$M = N$$.

In other words, if a square matrix $$A$$ has a left inverse $$M$$ and a right inverse $$N$$, then $$M$$ and $$N$$ must be the same matrix.

To see this, multiply both sides of $$MA=I$$ on the right by $$N$$ to get $$(MA)N = N$$. But $(MA)N = M(AN) = MI = M.$ Hence, $$M = N$$.

If $$B$$ is a matrix such that $$BA = AB = I$$, $$B$$ is said to be an inverse matrix of $$A$$. One can easily show that inverse matrices are unique. The inverse matrix of $$A$$ is written as $$A^{-1}$$.

### Example

Let $$A = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & -2 \\ 0 & 1\end{bmatrix}$$. One can easily check that $$AB = BA = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$. Hence, $$B$$ is the inverse matrix of $$A$$ (and vice versa).

If a square matrix $$A$$ has an inverse, $$A$$ is said to be invertible.

## Exercise

Let $$A = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$$. Suppose that $$ad - bc \neq 0$$. Verify that $$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a\end{bmatrix}$$.