Inverse and Determinant

Let \(A\) be a square matrix of order \(n\). \(A\) is invertible if and only if \(\text{det}(A)\neq0\).

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Invertible \(\implies\) Non-zero determinant

If \(A\) is invertible, \(A^{-1}\) exists. We have:

\[ \begin{aligned} AA^{-1} & =I\\ \\\text{det}(A)\cdot\text{det}\left(A^{-1}\right) & =1 \end{aligned} \]

Since the product of two real numbers is \(1\), both of them have to be non-zero, establishing that \(\text{det}(A)\neq0\). We can compute the determinant of \(A^{-1}\) and it is \(\cfrac{1}{\text{det}(A)}\).

Non-zero determinant \(\implies\) Invertible

We use the following identity:

\[ A \cdot \text{adj}(A)=\text{adj}(A)\cdot A=\text{det}(A)\cdot I \]

where \(\text{adj}(A)\) is the adjugate of \(A\). If \(\text{det}(A)\neq0\), we can divide the above identity uniformly by \(\text{det}(A)\) to get:

\[ \begin{aligned} A\cdot\left[\cfrac{1}{\text{det}(A)}\text{adj}(A)\right] & =\left[\cfrac{1}{\text{det}(A)}\text{adj}(A)\right]A=I \end{aligned} \]

We now have a matrix \(B\) such that \(AB=BA=I\). It follows that \(A\) is invertible. Specifically, the inverse of \(A\) can be represented as:

\[ A^{-1}=\cfrac{1}{\text{det}(A)}\text{adj}(A) \]