Let \(A\) be a \(2 \times 2\) matrix over a field \(F\). Show that \(A\) is invertible if and only if \(\det A \neq 0\). When \(A\) is invertible, give a formula for \(A^{-1}\).
Solution
Write \[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}. \] If \(\det A=ad-bc\ne0\), then the adjoint identity gives \[ A\begin{bmatrix}d&-b\\-c&a\end{bmatrix} =(ad-bc)I, \] so \[ A^{-1}=\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}. \] Thus \(\det A\ne0\) implies that \(A\) is invertible.
Conversely, suppose \(\det A=0\). If \((a,b)\ne(0,0)\), then \[ x=(-b,a)^t\ne0 \] and \[ Ax= \begin{bmatrix} -ab+ab\\ -bc+ad \end{bmatrix} =0. \] Hence \(A\) has a nonzero kernel vector and is not invertible. If \((a,b)=(0,0)\), then the first row of \(A\) is zero, so again \(A\) is not invertible. Therefore invertibility implies \(\det A\ne0\).