1.4 - Inverting matrix products

1.4 - Inverting matrix products

Reversing a sequence requires undoing its last step first. Derive the inverse-product rule, extend it to longer products, and use it to rearrange matrix equations while keeping every factor in its proper position.

Reversing the order

If M=AB\mathbf M=\mathbf{AB}, should its inverse be A−1B−1\mathbf A^{-1}\mathbf B^{-1} or B−1A−1\mathbf B^{-1}\mathbf A^{-1}? Test which order places inverse pairs next to each other.

For example, A=(1101)\mathbf A=\begin{pmatrix}1&1\\0&1\end{pmatrix} has inverse (1−101)\begin{pmatrix}1&-1\\0&1\end{pmatrix}, and B=(2001)\mathbf B=\begin{pmatrix}2&0\\0&1\end{pmatrix} has inverse (12001)\begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix}. The product AB\mathbf{AB} first multiplies a column by B\mathbf B, then by A\mathbf A. Recovery must therefore use A−1\mathbf A^{-1} first, then B−1\mathbf B^{-1}; the matrix for that recovery is written B−1A−1\mathbf B^{-1}\mathbf A^{-1}.

Inverse of a product

For non-singular square matrices A,B\mathbf A,\mathbf B of the same size,

(AB)−1=B−1A−1.(\mathbf{AB})^{-1}=\mathbf B^{-1}\mathbf A^{-1}.

Reverse the factor order and invert every factor. Not in the formula booklet: learn it.

The algebra proves both sides of the inverse condition without commuting any factors:

(AB)(B−1A−1)=A(BB−1)A−1=I(B−1A−1)(AB)=B−1(A−1A)B=I.\begin{aligned} (\mathbf{AB})(\mathbf B^{-1}\mathbf A^{-1}) &=\mathbf A(\mathbf B\mathbf B^{-1})\mathbf A^{-1}=\mathbf I\\ (\mathbf B^{-1}\mathbf A^{-1})(\mathbf{AB}) &=\mathbf B^{-1}(\mathbf A^{-1}\mathbf A)\mathbf B=\mathbf I. \end{aligned}

Only associativity and adjacent inverse pairs are used. This rule assumes every stated inverse exists.

Worked example 1: invert the factors, then reverse

Question. For A=(1101)\mathbf A=\begin{pmatrix}1&1\\0&1\end{pmatrix}, B=(2001)\mathbf B=\begin{pmatrix}2&0\\0&1\end{pmatrix} and M=AB\mathbf M=\mathbf{AB}, express M−1\mathbf M^{-1} as a product of two non-identity matrices and then evaluate it.

Plan. Write B−1A−1\mathbf B^{-1}\mathbf A^{-1}, keeping the requested product form visible before multiplying.

M−1=(12001)(1−101)=(12−1201).\begin{aligned} \mathbf M^{-1} &=\begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix}\\ &=\begin{pmatrix}\tfrac12&-\tfrac12\\0&1\end{pmatrix}. \end{aligned}

Answer. The first line is the required product; its value is (12−1201)\begin{pmatrix}\tfrac12&-\tfrac12\\0&1\end{pmatrix}.

Check. Direct multiplication first gives M=(2101)\mathbf M=\begin{pmatrix}2&1\\0&1\end{pmatrix}. Its two-by-two inverse formula gives 12(1−102)\tfrac12\begin{pmatrix}1&-1\\0&2\end{pmatrix}, agreeing with the independently constructed answer.

Change only the order in the original product. The inverses now appear in the other order too.

Worked example 2: swapping the original order

Question. For the same matrices, let N=BA\mathbf N=\mathbf{BA}. Find N−1\mathbf N^{-1} as a product and as a single matrix.

Plan. Reverse the order in BA\mathbf{BA}, giving A−1B−1\mathbf A^{-1}\mathbf B^{-1}.

N−1=(1−101)(12001)=(12−101).\mathbf N^{-1} =\begin{pmatrix}1&-1\\0&1\end{pmatrix} \begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix} =\begin{pmatrix}\tfrac12&-1\\0&1\end{pmatrix}.

Answer. N−1=A−1B−1=(12−101)\mathbf N^{-1}=\mathbf A^{-1}\mathbf B^{-1}=\begin{pmatrix}\tfrac12&-1\\0&1\end{pmatrix}.

Check. N=(2201)\mathbf N=\begin{pmatrix}2&2\\0&1\end{pmatrix}, and NN−1=(1−2+201)=I\mathbf N\mathbf N^{-1}=\begin{pmatrix}1&-2+2\\0&1\end{pmatrix}=\mathbf I. Using the preceding example's inverse instead would leave a nonzero top-right entry.

The two final matrices differ. Inverting each factor without reversing their order would therefore answer a different question.

Exam tip

If asked to write an inverse as a product, show the inverse factors in the required order. A single multiplied-out matrix does not display that requested form. Retain each determinant divisor in its factor.

Keep a product form in the next answer.

More than two factors

A longer product is reversed by undoing every factor in the opposite order, not by reversing only one neighbouring pair.

For three non-singular matrices, regard ABC\mathbf{ABC} as (AB)C(\mathbf{AB})\mathbf C. The two-factor rule gives

((AB)C)−1=C−1(AB)−1=C−1B−1A−1.\bigl((\mathbf{AB})\mathbf C\bigr)^{-1} =\mathbf C^{-1}(\mathbf{AB})^{-1} =\mathbf C^{-1}\mathbf B^{-1}\mathbf A^{-1}.

This grouping explains the extension rather than introducing a new unrelated rule.

Inverse of a longer product

For non-singular matrices of the same size,

(A1A2⋯An)−1=An−1⋯A2−1A1−1.(\mathbf A_1\mathbf A_2\cdots\mathbf A_n)^{-1} =\mathbf A_n^{-1}\cdots\mathbf A_2^{-1}\mathbf A_1^{-1}.

In particular, (An)−1=(A−1)n(\mathbf A^n)^{-1}=(\mathbf A^{-1})^n for positive integer nn. Not in the formula booklet: learn it.

Repeated factors still count separately. In ABA\mathbf{ABA}, reverse the list A,B,A\mathbf A,\mathbf B,\mathbf A to obtain A−1B−1A−1\mathbf A^{-1}\mathbf B^{-1}\mathbf A^{-1}, not A−2B−1\mathbf A^{-2}\mathbf B^{-1} unless the factors commute.

Worked example 3: three factors

Question. Let

A=(1101),B=(2001),C=(1011).\mathbf A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\quad \mathbf B=\begin{pmatrix}2&0\\0&1\end{pmatrix},\quad \mathbf C=\begin{pmatrix}1&0\\1&1\end{pmatrix}.

Find (ABC)−1(\mathbf{ABC})^{-1}.

Plan. Invert each factor and multiply in the order C−1B−1A−1\mathbf C^{-1}\mathbf B^{-1}\mathbf A^{-1}.

Here C−1=(10−11)\mathbf C^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}. Thus

(ABC)−1=(10−11)(12001)(1−101)=(120−121)(1−101)=(12−12−1232).\begin{aligned} (\mathbf{ABC})^{-1} &=\begin{pmatrix}1&0\\-1&1\end{pmatrix} \begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix}\\ &=\begin{pmatrix}\tfrac12&0\\-\tfrac12&1\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix}\\ &=\begin{pmatrix}\tfrac12&-\tfrac12\\-\tfrac12&\tfrac32\end{pmatrix}. \end{aligned}

Answer. 12(1−1−13)\tfrac12\begin{pmatrix}1&-1\\-1&3\end{pmatrix}.

Check. Directly multiplying the original factors gives ABC=(3111)\mathbf{ABC}=\begin{pmatrix}3&1\\1&1\end{pmatrix}, whose determinant is 22. The two-by-two inverse formula gives the same answer by a different route.

Swap only the first two factors in the original list. Keep the third factor fixed, then reverse the whole new list.

Worked example 4: a changed internal order

Question. With the same matrices, find (BAC)−1(\mathbf{BAC})^{-1}.

Plan. Reverse B,A,C\mathbf B,\mathbf A,\mathbf C to obtain C−1A−1B−1\mathbf C^{-1}\mathbf A^{-1}\mathbf B^{-1}.

(BAC)−1=(10−11)(1−101)(12001)=(1−1−12)(12001)=(12−1−122).\begin{aligned} (\mathbf{BAC})^{-1} &=\begin{pmatrix}1&0\\-1&1\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix} \begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix}\\ &=\begin{pmatrix}1&-1\\-1&2\end{pmatrix} \begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix}\\ &=\begin{pmatrix}\tfrac12&-1\\-\tfrac12&2\end{pmatrix}. \end{aligned}

Answer. 12(1−2−14)\tfrac12\begin{pmatrix}1&-2\\-1&4\end{pmatrix}.

Check. BAC=(4211)\mathbf{BAC}=\begin{pmatrix}4&2\\1&1\end{pmatrix}. Multiplying it by the answer gives (2−1−4+412−12−1+2)=I\begin{pmatrix}2-1&-4+4\\\tfrac12-\tfrac12&-1+2\end{pmatrix}=\mathbf I.

A symbolic check is also useful: in ABCC−1B−1A−1\mathbf{ABC}\mathbf C^{-1}\mathbf B^{-1}\mathbf A^{-1}, the central CC−1\mathbf C\mathbf C^{-1} cancels first, then the B\mathbf B pair, then the A\mathbf A pair. No factor has to cross another.

Watch out

Treat (A−1)−1=A(\mathbf A^{-1})^{-1}=\mathbf A as an inverse of a whole factor. For example, (AB−1C)−1=C−1BA−1(\mathbf A\mathbf B^{-1}\mathbf C)^{-1}=\mathbf C^{-1}\mathbf B\mathbf A^{-1}.

Reverse a list that contains an inverse factor.

Equations with products

When known matrices surround an unknown, remove the left factor from the left and the right factor from the right.

For non-singular A,B\mathbf A,\mathbf B, the equation AXB=C\mathbf{AXB}=\mathbf C gives

A−1(AXB)=A−1CXB=A−1CX=A−1CB−1.\begin{aligned} \mathbf A^{-1}(\mathbf{AXB})&=\mathbf A^{-1}\mathbf C\\ \mathbf{XB}&=\mathbf A^{-1}\mathbf C\\ \mathbf X&=\mathbf A^{-1}\mathbf C\mathbf B^{-1}. \end{aligned}

You could remove B\mathbf B first instead; the same final order follows. The unknown need not be invertible for this rearrangement, because only A\mathbf A and B\mathbf B are cancelled.

Worked example 5: factors on both sides

Question. Solve AXB=C\mathbf{AXB}=\mathbf C, where

A=(1101),B=(2001),C=(24−23).\mathbf A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\quad \mathbf B=\begin{pmatrix}2&0\\0&1\end{pmatrix},\quad \mathbf C=\begin{pmatrix}2&4\\-2&3\end{pmatrix}.

Plan. Place A−1\mathbf A^{-1} on the left and B−1\mathbf B^{-1} on the right.

X=(1−101)(24−23)(12001)=(41−23)(12001)=(21−13).\begin{aligned} \mathbf X &=\begin{pmatrix}1&-1\\0&1\end{pmatrix} \begin{pmatrix}2&4\\-2&3\end{pmatrix} \begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix}\\ &=\begin{pmatrix}4&1\\-2&3\end{pmatrix} \begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix} =\begin{pmatrix}2&1\\-1&3\end{pmatrix}. \end{aligned}

Answer. X=(21−13)\mathbf X=\begin{pmatrix}2&1\\-1&3\end{pmatrix}.

Check. First XB=(41−23)\mathbf{XB}=\begin{pmatrix}4&1\\-2&3\end{pmatrix}; then A(XB)=(24−23)=C\mathbf A(\mathbf{XB})=\begin{pmatrix}2&4\\-2&3\end{pmatrix}=\mathbf C. This substitutes into the original equation.

Change only which known matrix stands on each side of the unknown. Keep the right-hand side unchanged; a different unknown matrix is needed.

Worked example 6: exchanging the surrounding factors

Question. For the same A,B,C\mathbf A,\mathbf B,\mathbf C, solve BXA=C\mathbf{BXA}=\mathbf C.

Plan. Cancel B\mathbf B from the left and A\mathbf A from the right.

X=B−1CA−1=(12001)(24−23)(1−101)=(12−23)(1−101)=(11−25).\begin{aligned} \mathbf X &=\mathbf B^{-1}\mathbf C\mathbf A^{-1}\\ &=\begin{pmatrix}\tfrac12&0\\0&1\end{pmatrix} \begin{pmatrix}2&4\\-2&3\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix}\\ &=\begin{pmatrix}1&2\\-2&3\end{pmatrix} \begin{pmatrix}1&-1\\0&1\end{pmatrix} =\begin{pmatrix}1&1\\-2&5\end{pmatrix}. \end{aligned}

Answer. X=(11−25)\mathbf X=\begin{pmatrix}1&1\\-2&5\end{pmatrix}.

Check. XA=(12−23)\mathbf{XA}=\begin{pmatrix}1&2\\-2&3\end{pmatrix}, and multiplying by B\mathbf B on the left gives C\mathbf C, as required.

For an equation such as ABX=D\mathbf{ABX}=\mathbf D, the left factor is the whole product AB\mathbf{AB}. Its inverse is B−1A−1\mathbf B^{-1}\mathbf A^{-1}, so X=B−1A−1D\mathbf X=\mathbf B^{-1}\mathbf A^{-1}\mathbf D. This returns to the opening question: inverse order follows from the cancellation you need, not from a rule that factors can be rearranged freely.

Problem-solving

Write one cancellation line before multiplying numerical matrices. A correctly ordered expression for the unknown is easier to check than a guessed final array.

Apply the same two-sided cancellation with new entries.

Quick Recap

  • Invert every factor and reverse the whole order.
  • For three factors, (ABC)−1=C−1B−1A−1(\mathbf{ABC})^{-1}=\mathbf C^{-1}\mathbf B^{-1}\mathbf A^{-1}.
  • Every inverse used must exist; repeated factors and already inverted factors still need careful treatment.
  • Remove surrounding factors from their own sides, then check the original equation.

The inverse of AB\mathbf{AB} is B−1A−1\mathbf B^{-1}\mathbf A^{-1}: this is the order that lets adjacent inverse pairs cancel.