The Inverse of a Matrix Product

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, with a formal proof and a fully worked example.

The Theorem

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.

Proof

To prove the theorem, we must show that \( (AB)(B^{-1}A^{-1}) = I \) and \( (B^{-1}A^{-1})(AB) = I \), where \( I \) is the identity matrix.

First direction:

\[ \begin{aligned} (AB)(B^{-1}A^{-1}) &= A (B B^{-1}) A^{-1} \\ &= A (I) A^{-1} \\ &= A A^{-1} \\ &= I \end{aligned} \]

Second direction:

\[ \begin{aligned} (B^{-1}A^{-1})(AB) &= B^{-1} (A^{-1} A) B \\ &= B^{-1} (I) B \\ &= B^{-1} B \\ &= I \end{aligned} \]

Since \( (AB)(B^{-1}A^{-1}) = (B^{-1}A^{-1})(AB) = I \), the matrix \( B^{-1}A^{-1} \) is indeed the inverse of \( AB \). Hence \( (AB)^{-1} = B^{-1}A^{-1} \).

Example with 2 by 2 Matrices

Let us verify the theorem on two \(2 \times 2\) matrices:

\[ A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \]

Step 1 — Compute the product \(AB\).

\[ \begin{aligned} AB &= \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \\ &= \begin{pmatrix} 2(1) + 1(3) & 2(2) + 1(4) \\ 1(1) + 1(3) & 1(2) + 1(4) \end{pmatrix} \\ &= \begin{pmatrix} 5 & 8 \\ 4 & 6 \end{pmatrix} \end{aligned} \]

Step 2 — Find the inverse of \(AB\) directly.

For a \(2 \times 2\) 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 \( AB = \begin{pmatrix} 5 & 8 \\ 4 & 6 \end{pmatrix} \), the determinant is \( 5 \times 6 - 8 \times 4 = 30 - 32 = -2 \). Hence

\[ (AB)^{-1} = \frac{1}{-2}\begin{pmatrix} 6 & -8 \\ -4 & 5 \end{pmatrix} = \begin{pmatrix} -3 & 4 \\ 2 & -\tfrac{5}{2} \end{pmatrix} \]

Step 3 — Compute \(A^{-1}\) and \(B^{-1}\) separately.

For \( A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \), \( \det(A) = 2 \times 1 - 1 \times 1 = 1 \), so

\[ A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \]

For \( B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), \( \det(B) = 1 \times 4 - 2 \times 3 = -2 \), so

\[ B^{-1} = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \tfrac{3}{2} & -\tfrac{1}{2} \end{pmatrix} \]

Step 4 — Compute \(B^{-1}A^{-1}\).

\[ \begin{aligned} B^{-1}A^{-1} &= \begin{pmatrix} -2 & 1 \\ \tfrac{3}{2} & -\tfrac{1}{2} \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} \\ &= \begin{pmatrix} (-2)(1) + (1)(-1) & (-2)(-1) + (1)(2) \\ (\tfrac{3}{2})(1) + (-\tfrac{1}{2})(-1) & (\tfrac{3}{2})(-1) + (-\tfrac{1}{2})(2) \end{pmatrix} \\ &= \begin{pmatrix} -3 & 4 \\ 2 & -\tfrac{5}{2} \end{pmatrix} \end{aligned} \]

Conclusion: \( (AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} -3 & 4 \\ 2 & -\tfrac{5}{2} \end{pmatrix} \), which confirms the theorem.

Important Notes

  • Order reversal is crucial: \( (AB)^{-1} = B^{-1}A^{-1} \), not \( A^{-1}B^{-1} \).
  • Extension to more matrices: \( (ABC)^{-1} = C^{-1}B^{-1}A^{-1} \).
  • Validity condition: All matrices involved must be invertible (square matrices with nonzero determinants).
  • Non-commutativity: In general, \( A^{-1}B^{-1} \neq B^{-1}A^{-1} \) unless \( A \) and \( B \) commute.

Applications

This property is useful in many areas of mathematics and its applications:

References