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## 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**.

Quick Quiz

## Exercise

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