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Chapter 4 · Determinants

Example 14: Verify (AB)−1=B−1A−1(AB)^{-1}=B^{-1}A^{-1}

ExamplesInverse of a matrix · Section 4.5
The question

If

A=[231−4]andB=[1−2−13],A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \quad\text{and}\quad B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix},

then verify that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}⁠.

Exam answer AB=[231−4][1−2−13]=[2−3−4+91+4−2−12]=[−155−14]\begin{aligned} AB &= \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \\ &= \begin{bmatrix} 2-3 & -4+9 \\ 1+4 & -2-12 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix} \end{aligned}

∣AB∣=14−25=−11≠0|AB| = 14 - 25 = -11 \ne 0⁠, so

(AB)−1=1∣AB∣adj⁡(AB)=−111[−14−5−5−1]=111[14551]…(i)\begin{aligned} (AB)^{-1} &= \frac{1}{|AB|}\operatorname{adj}(AB) \\ &= -\frac{1}{11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} \\ &= \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}\quad\ldots(\text{i}) \end{aligned}

∣A∣=−8−3=−11≠0|A| = -8 - 3 = -11 \ne 0 and ∣B∣=3−2=1≠0|B| = 3 - 2 = 1 \ne 0⁠, so

A−1=1∣A∣adj⁡A=−111[−4−3−12]B−1=1∣B∣adj⁡B=[3211]\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= -\frac{1}{11}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} \\ B^{-1} &= \frac{1}{|B|}\operatorname{adj}B \\ &= \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \end{aligned} B−1A−1=−111[3211][−4−3−12]=−111[−12−2−9+4−4−1−3+2]=111[14551]…(ii)\begin{aligned} &B^{-1}A^{-1} \\ &= -\frac{1}{11}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} \\ &= -\frac{1}{11}\begin{bmatrix} -12-2 & -9+4 \\ -4-1 & -3+2 \end{bmatrix} \\ &= \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}\quad\ldots(\text{ii}) \end{aligned}

From (i) and (ii), (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}⁠. Hence verified.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

To undo "first BB⁠, then AA" you must undo AA first and BB second, like taking off shoes before socks. So the inverse of a product is the product of the inverses in reverse order. Here we confirm it with numbers: compute (AB)−1(AB)^{-1} directly, compute B−1A−1B^{-1}A^{-1} separately, and see that they are the same matrix.

Given, to show, and the plan
Given
Two 2×22\times2 matrices AA and BB⁠.
To show
(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} for these matrices.
Plan
(1) Compute ABAB⁠, its determinant and its inverse. (2) Compute A−1A^{-1} and B−1B^{-1} separately. (3) Multiply B−1A−1B^{-1}A^{-1} and compare. (4) See why the order must reverse, and check that the other order fails.

Basics you need

1Inverse of a 2×22\times2 matrix

For MM with entries a,ba, b (top row) and c,dc, d (bottom row): ∣M∣=ad−bc|M| = ad - bc⁠. If ∣M∣≠0|M| \ne 0⁠, then M−1=1∣M∣adj⁡MM^{-1} = \frac{1}{|M|}\operatorname{adj}M⁠, where adj⁡M\operatorname{adj}M swaps aa and dd and changes the signs of bb and cc⁠. Basics: why the inverse divides by |A|

2Matrix multiplication

Entry (i,j)(i,j) of PQPQ is row ii of PP times column jj of QQ⁠. In general PQ≠QPPQ \ne QP⁠, so the order of factors matters. Basics: matrix facts the chapter uses

3Scalars move freely

A number factor can be taken out of a product: (kP)(mQ)=km (PQ)(kP)(mQ) = km\,(PQ)⁠. Basics: scaling and adding matrices

Solution

  1. Compute ABAB

    AA and BB are both 2×22\times2⁠: columns of AA (2) = rows of BB (2), so ABAB is defined and is 2×22\times2⁠. Row ii of AA times column jj of BB⁠:

    (AB)11=2⋅1+3(−1)=−1,(AB)12=2(−2)+3⋅3=5,(AB)21=1⋅1+(−4)(−1)=5,(AB)22=1(−2)+(−4)⋅3=−14.\begin{aligned} (AB)_{11} &= 2\cdot1 + 3(-1) = -1,\\ (AB)_{12} &= 2(-2) + 3\cdot3 = 5,\\ (AB)_{21} &= 1\cdot1 + (-4)(-1) = 5,\\ (AB)_{22} &= 1(-2) + (-4)\cdot3 = -14. \end{aligned} AB=[−155−14].AB = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}.
    The left side (AB)−1(AB)^{-1} is the inverse of this single matrix, so we need ABAB itself first. Every entry is written out because one sign slip here would spoil everything after it.
    Multiplication of matrices, Chapter 3, Section 3.4.5. Basics: matrix facts the chapter uses

  2. ∣AB∣|AB| and (AB)−1(AB)^{-1}

    The determinant:

    ∣AB∣=(−1)(−14)−5⋅5=14−25=−11.|AB| = (-1)(-14) - 5\cdot5 = 14 - 25 = -11.

    Since ∣AB∣=−11≠0|AB| = -11 \ne 0⁠, ABAB is non-singular and (AB)−1(AB)^{-1} exists. Its adjoint (swap −1-1 and −14-14⁠, change the signs of the two 5s):

    adj⁡(AB)=[−14−5−5−1].\operatorname{adj}(AB) = \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}.

    So

    (AB)−1=1∣AB∣adj⁡(AB)=−111[−14−5−5−1]=111[14551].\begin{aligned} (AB)^{-1} &= \frac{1}{|AB|}\operatorname{adj}(AB)\\ &= -\frac{1}{11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix}\\ &= \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}. \end{aligned}
    Theorem 4 needs ∣AB∣≠0|AB| \ne 0 before we may divide by it. The cofactors of the entries of ABAB are −14-14⁠, −5-5 (row 1) and −5-5⁠, −1-1 (row 2); their matrix is symmetric, so transposing it gives the same matrix, the adjoint above. In the last line, the outside minus sign was multiplied into every entry: −111⋅(−14)=1411-\frac{1}{11}\cdot(-14) = \frac{14}{11}⁠, and so on.
    Theorem 4, Section 4.5; the 2×22\times2 adjoint shortcut, Remark after Example 12. Lesson: inverse of a matrix

  3. A−1A^{-1}

    ∣A∣=2(−4)−3⋅1=−8−3=−11≠0|A| = 2(-4) - 3\cdot1 = -8 - 3 = -11 \ne 0⁠, so A−1A^{-1} exists. Swap 22 and −4-4⁠, change the signs of 33 and 11⁠:

    adj⁡A=[−4−3−12],\operatorname{adj}A = \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}, A−1=−111[−4−3−12].A^{-1} = -\frac{1}{11}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}.
    The right side B−1A−1B^{-1}A^{-1} needs both inverses separately, and each needs its own non-zero determinant. We keep the factor −111-\frac{1}{11} outside, as the textbook does, because it makes the multiplication in Step 5 easier.
    A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A when ∣A∣≠0|A| \ne 0⁠. Basics: why the inverse divides by |A|

  4. B−1B^{-1}

    ∣B∣=1⋅3−(−2)(−1)=3−2=1≠0|B| = 1\cdot3 - (-2)(-1) = 3 - 2 = 1 \ne 0⁠, so B−1B^{-1} exists. Swap 11 and 33⁠, change the signs of −2-2 and −1-1⁠:

    adj⁡B=[3211],\operatorname{adj}B = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}, B−1=11adj⁡B=[3211].B^{-1} = \frac{1}{1}\operatorname{adj}B = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}.
    Changing the sign of a negative entry makes it positive: −(−2)=2-(-2) = 2 and −(−1)=1-(-1) = 1⁠. With ∣B∣=1|B| = 1⁠, the inverse equals the adjoint.

  5. Multiply B−1A−1B^{-1}A^{-1}

    Take the scalar −111-\frac{1}{11} outside and multiply the two integer matrices, row of the first times column of the second:

    B−1A−1=−111[3211][−4−3−12].B^{-1}A^{-1} = -\frac{1}{11}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}. (1,1): 3(−4)+2(−1)=−14,(1,2): 3(−3)+2⋅2=−5,(2,1): 1(−4)+1(−1)=−5,(2,2): 1(−3)+1⋅2=−1.\begin{aligned} (1,1)&: \ 3(-4) + 2(-1) = -14,\\ (1,2)&: \ 3(-3) + 2\cdot2 = -5,\\ (2,1)&: \ 1(-4) + 1(-1) = -5,\\ (2,2)&: \ 1(-3) + 1\cdot2 = -1. \end{aligned} B−1A−1=−111[−14−5−5−1]=111[14551].B^{-1}A^{-1} = -\frac{1}{11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}.

    This is exactly the matrix found in Step 2. Hence (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

    Moving the number −111-\frac{1}{11} to the front is allowed because scalars commute with matrices. Then the two answers are compared entry by entry: 1411,511,511,111\frac{14}{11}, \frac{5}{11}, \frac{5}{11}, \frac{1}{11} in both.
    (kP)Q=k(PQ)(kP)Q = k(PQ)⁠: each entry of (kP)Q(kP)Q is a sum of products that all carry the factor kk⁠, so kk can be taken out of the sum (row-times-column rule, Chapter 3, Section 3.4.5). Basics: scaling and adding matrices

  6. Why the order reverses

    For any invertible AA⁠, BB of the same order, multiply ABAB by B−1A−1B^{-1}A^{-1} and regroup (multiplication is associative):

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

    In the same way

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

    So B−1A−1B^{-1}A^{-1} is an inverse of ABAB⁠, and since an inverse is unique, (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}⁠.

    The other order does not work. Here

    A−1B−1=−111[−4−3−12][3211],A^{-1}B^{-1} = -\frac{1}{11}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}, (1,1): (−4)⋅3+(−3)⋅1=−15,(1,2): (−4)⋅2+(−3)⋅1=−11,(2,1): (−1)⋅3+2⋅1=−1,(2,2): (−1)⋅2+2⋅1=0,\begin{aligned} (1,1)&: \ (-4)\cdot3 + (-3)\cdot1 = -15,\\ (1,2)&: \ (-4)\cdot2 + (-3)\cdot1 = -11,\\ (2,1)&: \ (-1)\cdot3 + 2\cdot1 = -1,\\ (2,2)&: \ (-1)\cdot2 + 2\cdot1 = 0, \end{aligned} A−1B−1=−111[−15−11−10],A^{-1}B^{-1} = -\frac{1}{11}\begin{bmatrix} -15 & -11 \\ -1 & 0 \end{bmatrix},

    which differs from (AB)−1(AB)^{-1} (its (2,2)(2,2) entry is 0, not 111\frac{1}{11}⁠).

    The numerical check of Step 5 confirms one case; the four-line argument shows it is true for every pair of invertible matrices of the same order. The regrouping collapses only because BB meets B−1B^{-1} in the middle, and that needs the reversed order. Because matrix multiplication does not commute, A−1B−1A^{-1}B^{-1} is a genuinely different matrix.
    Theorem 3 (uniqueness of inverse) and Theorem 4 ((AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}⁠), Chapter 3, Section 3.7. Lesson: inverse of a matrix
Result AB=[−155−14],∣AB∣=−11,AB = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}, \qquad |AB| = -11, (AB)−1=111[14551]=B−1A−1,(AB)^{-1} = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} = B^{-1}A^{-1},

with A−1=−111adj⁡AA^{-1} = -\frac{1}{11}\operatorname{adj}A (where ∣A∣=−11|A| = -11⁠) and B−1=adj⁡BB^{-1} = \operatorname{adj}B (where ∣B∣=1|B| = 1⁠). Hence (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} is verified. (This agrees with the textbook's own solution.)

Check it with numbers

Multiply ABAB by the inverse found in Step 2; the identity must appear:

[−155−14]⋅111[14551]=111[110011].\begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}\cdot\frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 11 & 0 \\ 0 & 11 \end{bmatrix}.

Entry by entry: −14+25=11-14 + 25 = 11⁠, −5+5=0-5 + 5 = 0⁠, 70−70=070 - 70 = 0⁠, 25−14=1125 - 14 = 11⁠. So the product is II⁠. Also ∣AB∣=−11=(−11)(1)=∣A∣ ∣B∣|AB| = -11 = (-11)(1) = |A|\,|B|⁠, as Theorem 3 of Section 4.5 says it must be.

Common mistakes
  1. Writing (AB)−1=A−1B−1(AB)^{-1} = A^{-1}B^{-1}⁠. Step 6 shows this gives a different matrix. The order reverses.
  2. Losing the minus sign in 1∣A∣\frac{1}{|A|}⁠. ∣A∣=−11|A| = -11⁠, so the factor is −111-\frac{1}{11}⁠.
  3. Sign errors in the 2×22\times2 adjoint. Only the off-diagonal entries change sign: in BB⁠, −2-2 and −1-1 become 22 and 11⁠; the diagonal 11 and 33 only swap.
  4. Skipping the determinant check. An inverse exists only when the determinant is non-zero; check ∣AB∣|AB|⁠, ∣A∣|A| and ∣B∣|B| before dividing.
What to take away
  1. (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}⁠: reverse the order, because BB must meet B−1B^{-1} in the middle.
  2. For 2×22\times2 matrices: M−1=1∣M∣adj⁡MM^{-1} = \frac{1}{|M|}\operatorname{adj}M⁠, with the swap-and-change-sign adjoint.
  3. ∣AB∣=∣A∣ ∣B∣|AB| = |A|\,|B|⁠, so ABAB is invertible exactly when both AA and BB are.

Every fact used, and where it comes from

FactUsed inSource
Product of matrices; order conditionSteps 1, 5, 6Chapter 3, Section 3.4.5 (Basics: matrix facts the chapter uses)
2×22\times2 determinant ad−bcad - bcSteps 2 to 4Section 4.2.2 (Lesson: determinant of a square matrix)
2×22\times2 adjoint: swap diagonal, change sign off itSteps 2 to 4Remark after Example 12, Section 4.5.1 (Lesson: the adjoint)
M−1=1∣M∣adj⁡MM^{-1} = \frac{1}{|M|}\operatorname{adj}M when ∣M∣≠0|M| \ne 0Steps 2 to 4Theorem 4, Section 4.5 (Lesson: inverse of a matrix)
Scalars factor out of productsStep 5Row-times-column rule, Chapter 3, Section 3.4.5
Associativity; uniqueness of inverse; (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}Step 6Chapter 3, Section 3.4.6; Theorems 3 and 4, Section 3.7
∣AB∣=∣A∣ ∣B∣|AB| = |A|\,|B|CheckTheorem 3, Section 4.5
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