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In regular arithmetic, you can undo multiplying by 55 simply by multiplying by its reciprocal, 15\frac{1}{5}, giving 5×15=15 \times \frac{1}{5} = 1.
Matrix division does not exist, but we can achieve the exact same undoing effect by finding a partner matrix called the multiplicative inverse, written as A−1A^{-1}.
For any 2×22 \times 2 matrix AA, its inverse A−1A^{-1} satisfies the property AA−1=A−1A=IA A^{-1} = A^{-1} A = I, where I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} is the identity matrix.
How do we construct this inverse matrix directly from the original four entries?

The 2×2 Inversion Formula

In 1858, British mathematician Arthur Cayley formally introduced matrix algebra and showed that inverting a 2×22 \times 2 matrix requires two quick actions: swapping one diagonal and changing the signs of the other.
For a general matrix A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, its adjugate matrix (or classical adjoint) is adj(A)=(d−b−ca)\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.
To complete the inverse, we divide the adjugate by the determinant of AA, defined as det⁡(A)=ad−bc\det(A) = ad - bc.
A−1=1det⁡(A)adj(A)=1ad−bc(d−b−ca)A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
A matrix is non-singular when det⁡(A)≠0\det(A) \neq 0, meaning the inverse exists; if det⁡(A)=0\det(A) = 0, we cannot divide by zero, so no inverse exists.
Why does this simple swap-and-negate recipe guarantee that multiplying produces the identity matrix?