In linear algebra, finding the inverse of a product of matrices is a common operation. This page explains the relationship between the inverse of a product and the inverses of the individual matrices.
If \(A\) and \(B\) are invertible matrices, then the product \(AB\) is also invertible, and:
\[(AB)^{-1} = B^{-1}A^{-1}\]
Note the order reversal: the inverse of a product is the product of the inverses in the reverse order.
To prove this theorem, we need to show that \((AB)(B^{-1}A^{-1}) = I\) and \((B^{-1}A^{-1})(AB) = I\), where \(I\) is the identity matrix.
Let's verify:
\begin{align} (AB)(B^{-1}A^{-1}) &= A(BB^{-1})A^{-1}\\ &= A(I)A^{-1}\\ &= AA^{-1}\\ &= I \end{align}
Similarly:
\begin{align} (B^{-1}A^{-1})(AB) &= B^{-1}(A^{-1}A)B\\ &= B^{-1}(I)B\\ &= B^{-1}B\\ &= I \end{align}
This proves that \(B^{-1}A^{-1}\) is indeed the inverse of \(AB\).
Let's work with two 2 by 2 matrices:
Matrix A: \[A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\]
Matrix B: \[B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\]
Step 1: Calculate the product \(AB\).
\begin{align} AB &= \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\\ &= \begin{pmatrix} (2 \times 1 + 1 \times 3) & (2 \times 2 + 1 \times 4) \\ (1 \times 1 + 1 \times 3) & (1 \times 2 + 1 \times 4) \end{pmatrix}\\ &= \begin{pmatrix} 5 & 8 \\ 4 & 6 \end{pmatrix} \end{align}
Step 2: Find the inverse of the product \(AB\) directly.
For a 2 by 2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the inverse is: \[\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\]
For our product matrix \(AB = \begin{pmatrix} 5 & 8 \\ 4 & 6 \end{pmatrix}\):
Determinant = \(5 \times 6 - 8 \times 4 = 30 - 32 = -2\)
\begin{align} (AB)^{-1} &= \frac{1}{-2}\begin{pmatrix} 6 & -8 \\ -4 & 5 \end{pmatrix}\\ &= \begin{pmatrix} -3 & 4 \\ 2 & -2.5 \end{pmatrix} \end{align}
Step 3: Calculate \(A^{-1}\) and \(B^{-1}\) separately.
For matrix \(A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\):
Determinant = \(2 \times 1 - 1 \times 1 = 1\)
\begin{align} A^{-1} &= \frac{1}{1}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\\ &= \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \end{align}
For matrix \(B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\):
Determinant = \(1 \times 4 - 2 \times 3 = 4 - 6 = -2\)
\begin{align} B^{-1} &= \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}\\ &= \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \end{align}
Step 4: Calculate \(B^{-1}A^{-1}\).
\begin{align} B^{-1}A^{-1} &= \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\\ &= \begin{pmatrix} (-2 \times 1 + 1 \times -1) & (-2 \times -1 + 1 \times 2) \\ (1.5 \times 1 + -0.5 \times -1) & (1.5 \times -1 + -0.5 \times 2) \end{pmatrix}\\ &= \begin{pmatrix} -3 & 4 \\ 2 & -2.5 \end{pmatrix} \end{align}
Verification: We can see that \((AB)^{-1} = B^{-1}A^{-1}\), which confirms the theorem.
This property is useful in many areas of mathematics and applications: