Miscellaneous · Q3: Find $(AB)^{-1}$ from $A^{-1}$ and $B$

Basics you need

Solution

1. Why $(AB)^{-1} = B^{-1}A^{-1}$

   A has an inverse (we are given it), so A is invertible. If B is also invertible (Step 2 shows $|B| \ne 0$⁠), then both are invertible $3\times3$ matrices and

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

   So $B^{-1}A^{-1}$ is the inverse of AB.

   A matrix that multiplies AB to give I on both sides is, by definition, the inverse of AB, and the inverse is unique. The two lines above use only the associative law and $BB^{-1} = I$⁠, $A^{-1}A = I$⁠. The order matters: $A^{-1}B^{-1}$ would leave $AB\,A^{-1}B^{-1}$⁠, in which B and $A^{-1}$ are not next to each other, so nothing cancels.
   Chapter 3, Theorem 4: $(AB)^{-1} = B^{-1}A^{-1}$⁠; also verified in Example 14. Lesson: inverse of a matrix

2. The determinant of B

   Row 3 of B is $(0, -2, 1)$ and contains a zero, so expand along $R_3$⁠, with signs $+, -, +$⁠:

   $$|B| = 0\cdot M_{31} - (-2)\,M_{32} + 1\cdot M_{33}.$$ $$M_{32} = \begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} = 0 - 2 = -2$$ $$M_{33} = \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 3 - (-2) = 5$$ $$|B| = 0 + 2(-2) + 5 = 1.$$

   Since $|B| = 1 \ne 0$⁠, B is non-singular and $B^{-1}$ exists.

   The zero in position $(3, 1)$ removes one term. In $M_{32}$ the products are $1\cdot0 = 0$ and $(-2)(-1) = 2$⁠, so $M_{32} = 0 - 2$⁠. In $M_{33}$ they are $1\cdot3 = 3$ and $2\cdot(-1) = -2$⁠. A square matrix is invertible exactly when its determinant is not zero.
   Section 4.5, Theorem 4: A is invertible if and only if A is non-singular. Lesson: expanding a 3 × 3 determinant

3. All nine cofactors of B

   The rows of B are $(1, 2, -2)$⁠, $(-1, 3, 0)$⁠, $(0, -2, 1)$⁠. For each position, delete its row and column, evaluate the $2\times2$ minor, and attach the sign $(-1)^{i+j}$⁠:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} = 3, & A_{11} &= 3 \\ M_{12} &= \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = -1, & A_{12} &= 1 \\ M_{13} &= \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} = 2, & A_{13} &= 2 \end{aligned}$$ $$\begin{aligned} M_{21} &= \begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} = -2, & A_{21} &= 2 \\ M_{22} &= \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1, & A_{22} &= 1 \\ M_{23} &= \begin{vmatrix} 1 & 2 \\ 0 & -2 \end{vmatrix} = -2, & A_{23} &= 2 \end{aligned}$$ $$\begin{aligned} M_{31} &= \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} = 6, & A_{31} &= 6 \\ M_{32} &= \begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} = -2, & A_{32} &= 2 \\ M_{33} &= \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 5, & A_{33} &= 5 \end{aligned}$$
   The positions with $i + j$ odd, namely $(1,2)$⁠, $(2,1)$⁠, $(2,3)$⁠, $(3,2)$⁠, change sign; the others keep it. For instance $M_{21} = 2\cdot1 - (-2)(-2)$ $= 2 - 4 = -2$⁠, so $A_{21} = (-1)^3(-2) = 2$⁠, and $M_{31} = 2\cdot0 - (-2)\cdot3 = 6$⁠. A quick consistency test: row 1 of B times its own cofactors must give $|B|$⁠: $1\cdot3 + 2\cdot1 + (-2)\cdot2 = 1$ ✓.
   Section 4.4, Definitions 1 and 2: $A_{ij} = (-1)^{i+j}M_{ij}$⁠. Lesson: minors and cofactors

4. adj B and $B^{-1}$

   Place the cofactors in their positions, then transpose:

   $$[A_{ij}] = \begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{bmatrix}, \quad \operatorname{adj}B = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}.$$

   Since $|B| = 1$⁠,

   $$B^{-1} = \frac{1}{1}\operatorname{adj}B = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}.$$
   The adjoint is the transpose of the cofactor matrix: row 1 of adj B is $(A_{11}, A_{21}, A_{31}) = (3, 2, 6)$⁠, the cofactors of column 1 of B. Forgetting the transpose is the most common error in this kind of question. Because $|B| = 1$⁠, dividing by $|B|$ changes nothing.
   Section 4.5.1, Definition 3 (adjoint); Theorem 4: $B^{-1} = \frac{1}{|B|}\operatorname{adj}B$⁠. Lesson: the adjoint

5. Check that $BB^{-1} = I$

   Rows of B: $(1, 2, -2)$⁠, $(-1, 3, 0)$⁠, $(0, -2, 1)$⁠. Columns of $B^{-1}$⁠: $(3, 1, 2)$⁠, $(2, 1, 2)$⁠, $(6, 2, 5)$⁠.

   $$\begin{aligned} (1,1)&: 3 + 2 - 4 = 1 \\ (1,2)&: 2 + 2 - 4 = 0 \\ (1,3)&: 6 + 4 - 10 = 0 \\ (2,1)&: -3 + 3 + 0 = 0 \\ (2,2)&: -2 + 3 + 0 = 1 \\ (2,3)&: -6 + 6 + 0 = 0 \\ (3,1)&: 0 - 2 + 2 = 0 \\ (3,2)&: 0 - 2 + 2 = 0 \\ (3,3)&: 0 - 4 + 5 = 1 \end{aligned}$$

   So $BB^{-1} = I$ ✓.

   A single slip in one cofactor would show up here as a wrong entry. For example, entry $(1,3)$ is $1\cdot6 + 2\cdot2 + (-2)\cdot5 = 6 + 4 - 10$⁠. Checking now prevents carrying an error into the final product.
   Product entry formula. Basics: matrix facts the chapter uses

6. Multiply $B^{-1}A^{-1}$

   Both factors are $3\times3$⁠, so the product is defined and is $3\times3$⁠. Rows of $B^{-1}$⁠: $(3, 2, 6)$⁠, $(1, 1, 2)$⁠, $(2, 2, 5)$⁠. Columns of $A^{-1}$⁠: $(3, -15, 5)$⁠, $(-1, 6, -2)$⁠, $(1, -5, 2)$⁠.

   $$\begin{aligned} c_{11} &= 9 - 30 + 30 = 9 \\ c_{12} &= -3 + 12 - 12 = -3 \\ c_{13} &= 3 - 10 + 12 = 5 \end{aligned}$$ $$\begin{aligned} c_{21} &= 3 - 15 + 10 = -2 \\ c_{22} &= -1 + 6 - 4 = 1 \\ c_{23} &= 1 - 5 + 4 = 0 \end{aligned}$$ $$\begin{aligned} c_{31} &= 6 - 30 + 25 = 1 \\ c_{32} &= -2 + 12 - 10 = 0 \\ c_{33} &= 2 - 10 + 10 = 2 \end{aligned}$$ $$(AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}.$$
   Each entry is a row of $B^{-1}$ times a column of $A^{-1}$⁠. In full, $c_{11} = 3\cdot3 + 2\cdot(-15) + 6\cdot5$ and $c_{31} = 2\cdot3 + 2\cdot(-15) + 5\cdot5$⁠; the other entries follow the same pattern, with the products listed in order.
   Product entry formula (Chapter 3, Section 3.4.5). Basics: matrix facts the chapter uses

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