Verify that (AB) inverse equals B inverse times A inverse

Problem

Verify that $(AB)^{-1} = B^{-1}A^{-1}$ with two $2\times 2$ matrices of your choice.

Solution

Let us choose $A = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ (both invertible, $\det A = 2$, $\det B = 1$). Left-hand side. $AB = \begin{pmatrix} 2 & 2 \\ 0 & 1\end{pmatrix}$, with $\det(AB) = 2$, therefore $(AB)^{-1} = \frac{1}{2}\begin{pmatrix} 1 & -2 \\ 0 & 2\end{pmatrix} = \begin{pmatrix} \tfrac{1}{2} & -1 \\ 0 & 1\end{pmatrix}.$ Right-hand side. $A^{-1} = \begin{pmatrix} \tfrac{1}{2} & 0 \\ 0 & 1\end{pmatrix}$, $\;B^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1\end{pmatrix}$, therefore $B^{-1}A^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1\end{pmatrix}\begin{pmatrix} \tfrac{1}{2} & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} \tfrac{1}{2} & -1 \\ 0 & 1\end{pmatrix}.$ The two sides coincide ✓. Note the reversed order of the factors: it is $B^{-1}A^{-1}$, not $A^{-1}B^{-1}$, because the product of matrices is not commutative.

Links

Topics: Linear systems
Concepts: Inverse matrix · Matrix product
Methods: 2x2 inverse matrix
Skills: Calculate · Prove
Exercise type: Proof