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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 \(A^{-1}\).


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


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