Exercise 4.4 · Q12: Verify $(AB)^{-1}=B^{-1}A^{-1}$

Basics you need

Solution

1. Compute $AB$

   $A$ is $2\times2$ and $B$ is $2\times2$⁠: columns of $A$ (2) = rows of $B$ (2), so $AB$ is defined and is $2\times2$⁠.

   $$\begin{aligned} c_{11} &= 3\cdot6 + 7\cdot7 = 18 + 49 = 67\\ c_{12} &= 3\cdot8 + 7\cdot9 = 24 + 63 = 87\\ c_{21} &= 2\cdot6 + 5\cdot7 = 12 + 35 = 47\\ c_{22} &= 2\cdot8 + 5\cdot9 = 16 + 45 = 61 \end{aligned}$$ $$AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$$
   $c_{12}$⁠, for example, pairs row 1 of $A$ ($3, 7$⁠) with column 2 of $B$ ($8, 9$⁠): $3\cdot8 + 7\cdot9$⁠. We need $AB$ as a single matrix because the left side asks for the inverse of the product.
   Pair up, multiply, add. Basics: matrix facts the chapter uses

2. The left side: $(AB)^{-1}$

   $$\begin{aligned} |AB| &= 67\cdot61 - 87\cdot47\\ &= 4087 - 4089 = -2 \end{aligned}$$

   $|AB| = -2 \neq 0$⁠, so $(AB)^{-1}$ exists. The cofactors of $AB$ are $A_{11} = 61$⁠, $A_{12} = -47$⁠, $A_{21} = -87$⁠, $A_{22} = 67$⁠, so

   $$\operatorname{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$$ $$\begin{aligned} (AB)^{-1} &= \frac{1}{-2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}\\[1mm] &= \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \end{aligned}$$
   For a $2\times2$ matrix each minor is a single entry: deleting row 1 and column 1 of $AB$ leaves $61$⁠, so $A_{11} = 61$⁠; deleting row 1 and column 2 leaves $47$⁠, and position $(1,2)$ carries a minus sign, so $A_{12} = -47$⁠; similarly $A_{21} = -87$ and $A_{22} = 67$⁠. Transposing $[A_{ij}]$ puts $-87$ in position $(1,2)$ and $-47$ in position $(2,1)$⁠. That is the "swap the diagonal, negate the rest" rule. (Here the cofactors $A_{ij}$ are those of the matrix $AB$⁠, not entries of $A$⁠.)
   $M^{-1} = \frac{1}{|M|}\operatorname{adj}M$⁠. Lesson: inverse of a matrix

3. $A^{-1}$

   $$|A| = \begin{vmatrix} 3 & 7 \\ 2 & 5 \end{vmatrix} = 3\cdot5 - 7\cdot2 = 1$$

   $|A| = 1 \neq 0$⁠, so $A^{-1}$ exists:

   $$\operatorname{adj}A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$ $$A^{-1} = \frac{1}{1}\operatorname{adj}A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$
   Swap the diagonal entries $3$ and $5$⁠; change the signs of $7$ and $2$⁠. Dividing by $|A| = 1$ changes nothing.

4. $B^{-1}$

   $$\begin{aligned} |B| &= \begin{vmatrix} 6 & 8 \\ 7 & 9 \end{vmatrix} = 6\cdot9 - 8\cdot7\\ &= 54 - 56 = -2 \end{aligned}$$

   $|B| = -2 \neq 0$⁠, so $B^{-1}$ exists:

   $$\operatorname{adj}B = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$$ $$B^{-1} = \frac{1}{-2}\operatorname{adj}B = -\frac{1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$$
   We keep the factor $-\frac12$ outside for now; it keeps the next multiplication in whole numbers. Notice that $|A|\,|B| = 1\cdot(-2) = -2 = |AB|$⁠, exactly as Theorem 3 ($|AB| = |A||B|$⁠) predicts: a first sign that the work is consistent.
   Lesson: the adjoint

5. The right side: $B^{-1}A^{-1}$

   Both are $2\times2$⁠, so the product is defined and is $2\times2$⁠. A scalar can be moved to the front of a product, so

   $$B^{-1}A^{-1} = -\frac{1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$

   Entries of the matrix product (rows of $\operatorname{adj}B$ times columns of $A^{-1}$⁠):

   $$\begin{aligned} d_{11} &= 9\cdot5 + (-8)(-2)\\ &= 45 + 16 = 61\\[1mm] d_{12} &= 9(-7) + (-8)(3)\\ &= -63 - 24 = -87\\[1mm] d_{21} &= (-7)(5) + 6(-2)\\ &= -35 - 12 = -47\\[1mm] d_{22} &= (-7)(-7) + 6\cdot3\\ &= 49 + 18 = 67 \end{aligned}$$ $$\begin{aligned} B^{-1}A^{-1} &= -\frac{1}{2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}\\[1mm] &= \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix} \end{aligned}$$

   Entry by entry this is the matrix found for $(AB)^{-1}$ in Step 2. Hence $(AB)^{-1} = B^{-1}A^{-1}$.

   Two matrices are equal when they have the same order and every pair of corresponding entries agrees. Both sides are $2\times2$ and all four entries match: $-\frac{61}{2}, \frac{87}{2}, \frac{47}{2}, -\frac{67}{2}$⁠. (In fact even before dividing, $\operatorname{adj}B\cdot\operatorname{adj}A$ equals $\operatorname{adj}(AB)$⁠.)
   $(kX)Y = k(XY)$ for a number $k$⁠. Basics: scaling and adding matrices

6. Why the order must reverse, for any invertible $A$⁠, $B$

   Let $A$ and $B$ be any invertible matrices of the same order. Then, using associativity and $BB^{-1} = I$⁠, $AA^{-1} = I$⁠:

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

   So $B^{-1}A^{-1}$ is an inverse of $AB$⁠, and since a matrix has only one inverse, it is the inverse: $(AB)^{-1} = B^{-1}A^{-1}$⁠.

   The numbers in Steps 1–5 verify the rule for one pair; this step shows why it always holds. In $B^{-1}A^{-1}$ the factor $B^{-1}$ is on the far left, so $A^{-1}$ meets $A$ in the middle first and cancels it, then $B^{-1}$ meets $B$⁠. With the order $A^{-1}B^{-1}$ nothing would sit next to its own inverse, because matrices do not commute in general.
   Uniqueness of the inverse and $(AB)^{-1} = B^{-1}A^{-1}$⁠: Theorems 3 and 4 of Chapter 3, Section 3.7. Basics: inverses

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