Inverses¶

Df

An Inverse of the matrix \(A\) is \(A^{-1}\) such that \(A^{-1}A=I\) where \(I\) is the Identity Matrix. Note that \(AA^{-1}=I\) also holds Inverses do not exist when there is a non-trivial vector \(\vec x\) such that \(A\vec x=\vec0\)

proof

Suppose \(A^{-1}\) exists in this case.
Then \(A^{-1}A\vec x=A^{-1}\vec 0\)
\(\Rightarrow I\vec x = \vec 0\) \(\Rightarrow \vec x = \vec 0\)

Finding Inverses¶

Gause-Jordan Method

Consider a matrix \(A\):

\[ A= \begin{bmatrix} 1&3\\ 2&7 \end{bmatrix} \]

and we want to find \(A^{-1}\) such that \(A^{-1}A=I\).
Suppose that \(A^{-1}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), then it is actually two equation sets:

\[ \begin{cases} \begin{bmatrix}a&b\end{bmatrix} \begin{bmatrix}1&3\\2&7\end{bmatrix} =\begin{bmatrix}1&0\end{bmatrix}\\ \begin{bmatrix}c&d\end{bmatrix} \begin{bmatrix}1&3\\2&7\end{bmatrix} =\begin{bmatrix}0&1\end{bmatrix} \end{cases} \]

One way to find \(A^{-1}\) is the Gause-Jordan method, and it is achieved first by adding the Identity Matrix to the right of \(A\) to form an Augmented Matrix \(\begin{bmatrix}1&3&1&0\\2&7&0&1\end{bmatrix}\). Then we perform elimination until the left side of the matrix is the Identity Matrix: \(\begin{bmatrix}1&0&7&-3\\0&1&-2&1\end{bmatrix}\), and the right side of the matrix, \(\begin{bmatrix}7&-3\\-2&1\end{bmatrix}\) is \(A^{-1}\). You can verify for yourself.

Proof that this is valid

Essentially we can rewrite the elimination process with a Elimination matrix, \(E\): what we are doing is essentially \(E\begin{bmatrix}A&I\end{bmatrix}\). Since the left side of the resulting product is \(I\), we know that \(EA=I\), and so \(E=A^{-1}\). Since the right side of the original augmented matrix is \(I\), the resulting right side is \(EI=E=A^{-1}\).