Miscellaneous · Q4: Verify $[\operatorname{adj}A]^{-1}=\operatorname{adj}(A^{-1})$ and $(A^{-1})^{-1}=A$

Basics you need

Part (i): $[\operatorname{adj}A]^{-1} = \operatorname{adj}(A^{-1})$

1. The determinant of A

   A has no zero entries, so we expand along $R_1 = (1, -2, 1)$⁠, with signs $+, -, +$⁠:

   $$M_{11} = \begin{vmatrix} 3 & 1 \\ 1 & 5 \end{vmatrix} = 15 - 1 = 14$$ $$M_{12} = \begin{vmatrix} -2 & 1 \\ 1 & 5 \end{vmatrix} = -10 - 1 = -11$$ $$M_{13} = \begin{vmatrix} -2 & 3 \\ 1 & 1 \end{vmatrix} = -2 - 3 = -5$$ $$\begin{aligned} |A| &= 1(14) - (-2)(-11) + 1(-5) \\ &= 14 - 22 - 5 = -13. \end{aligned}$$

   Since $|A| = -13 \ne 0$⁠, A is non-singular, so $A^{-1}$ exists.

   The middle term needs care: the pattern gives $-$⁠, the entry is $-2$⁠, and the minor is $-11$⁠. So it is $-(-2)(-11) = -22$⁠. The question only makes sense if $A^{-1}$ exists, which is exactly the condition $|A| \ne 0$⁠.
   Section 4.2.3 (expansion); Section 4.5, Theorem 4 (invertible ⇔ non-singular). Lesson: expanding a 3 × 3 determinant

2. All nine cofactors of A, and adj A

   Rows of A: $(1, -2, 1)$⁠, $(-2, 3, 1)$⁠, $(1, 1, 5)$⁠. The first row of minors comes from Part (i), Step 1; the rest:

   $$\begin{aligned} M_{21} &= \begin{vmatrix} -2 & 1 \\ 1 & 5 \end{vmatrix} = -11 \\ M_{22} &= \begin{vmatrix} 1 & 1 \\ 1 & 5 \end{vmatrix} = 4 \\ M_{23} &= \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = 3 \\ M_{31} &= \begin{vmatrix} -2 & 1 \\ 3 & 1 \end{vmatrix} = -5 \\ M_{32} &= \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = 3 \\ M_{33} &= \begin{vmatrix} 1 & -2 \\ -2 & 3 \end{vmatrix} = -1 \end{aligned}$$

   Now attach the signs $(-1)^{i+j}$⁠:

   $$\begin{aligned} A_{11} &= 14,\quad A_{12} = 11,\quad A_{13} = -5 \\ A_{21} &= 11,\quad A_{22} = 4,\quad A_{23} = -3 \\ A_{31} &= -5,\quad A_{32} = -3,\quad A_{33} = -1 \end{aligned}$$

   The matrix of cofactors is symmetric, so transposing it changes nothing:

   $$\operatorname{adj}A = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}.$$
   For example, $M_{23} = 1\cdot1 - (-2)\cdot1 = 3$ and $M_{33} = 1\cdot3 - (-2)(-2)$ $= 3 - 4 = -1$⁠. The signs change at the odd positions $(1,2)$⁠, $(2,1)$⁠, $(2,3)$⁠, $(3,2)$⁠. A itself is symmetric (entry $(i,j)$ equals entry $(j,i)$⁠), which is why its cofactor matrix is symmetric too; normally the transpose step does matter.
   Section 4.4, Definitions 1, 2; Section 4.5.1, Definition 3. Lesson: the adjoint

3. $A^{-1}$⁠, and the check $A\,(\operatorname{adj}A) = |A|\,I$

   Rows of A times columns of adj A, which are $(14, 11, -5)$⁠, $(11, 4, -3)$⁠, $(-5, -3, -1)$⁠:

   $$\begin{aligned} (1,1)&: 14 - 22 - 5 = -13 \\ (1,2)&: 11 - 8 - 3 = 0 \\ (1,3)&: -5 + 6 - 1 = 0 \\ (2,1)&: -28 + 33 - 5 = 0 \\ (2,2)&: -22 + 12 - 3 = -13 \\ (2,3)&: 10 - 9 - 1 = 0 \\ (3,1)&: 14 + 11 - 25 = 0 \\ (3,2)&: 11 + 4 - 15 = 0 \\ (3,3)&: -5 - 3 - 5 = -13 \end{aligned}$$

   So $A\,(\operatorname{adj}A) = -13\,I = |A|\,I$ ✓, and

   $$A^{-1} = \frac{1}{|A|}\operatorname{adj}A = -\frac{1}{13}\begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}.$$
   The check confirms all nine cofactors at once: a wrong cofactor would spoil at least one entry. Dividing $A\,(\operatorname{adj}A) = -13I$ by $-13$ gives $A \cdot \left(-\frac{1}{13}\operatorname{adj}A\right) = I$⁠, which is exactly the statement that $-\frac{1}{13}\operatorname{adj}A$ is the inverse. Keeping the factor $-\frac{1}{13}$ outside the matrix avoids fractions in every entry.
   Section 4.5, Theorem 1: $A\,(\operatorname{adj}A) = (\operatorname{adj}A)\,A = |A|\,I$⁠. Lesson: inverse of a matrix

4. The determinant of adj A

   Write $P = \operatorname{adj}A$⁠. Expand $|P|$ along its first row $(14, 11, -5)$⁠:

   $$\begin{vmatrix} 4 & -3 \\ -3 & -1 \end{vmatrix} = -4 - 9 = -13$$ $$\begin{vmatrix} 11 & -3 \\ -5 & -1 \end{vmatrix} = -11 - 15 = -26$$ $$\begin{vmatrix} 11 & 4 \\ -5 & -3 \end{vmatrix} = -33 + 20 = -13$$ $$\begin{aligned} |P| &= 14(-13) - 11(-26) + (-5)(-13) \\ &= -182 + 286 + 65 = 169. \end{aligned}$$

   So $|\operatorname{adj}A| = 169 \ne 0$⁠, and adj A is invertible.

   This agrees with the general fact $|\operatorname{adj}A| = |A|^{n-1}$ for an $n\times n$ matrix (the Remark after Theorem 3 in Section 4.5): here $(-13)^2 = 169$⁠. We still computed it directly, since the question asks us to verify, not to assume.
   Expansion along $R_1$⁠, Section 4.2.3. Lesson: expanding a 3 × 3 determinant

5. The cofactors of adj A, and $[\operatorname{adj}A]^{-1}$

   Rows of P: $(14, 11, -5)$⁠, $(11, 4, -3)$⁠, $(-5, -3, -1)$⁠. Row 1 of minors is in Part (i), Step 4; the others:

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 11 & -5 \\ -3 & -1 \end{vmatrix} = -26 \\ M_{22} &= \begin{vmatrix} 14 & -5 \\ -5 & -1 \end{vmatrix} = -39 \\ M_{23} &= \begin{vmatrix} 14 & 11 \\ -5 & -3 \end{vmatrix} = 13 \\ M_{31} &= \begin{vmatrix} 11 & -5 \\ 4 & -3 \end{vmatrix} = -13 \\ M_{32} &= \begin{vmatrix} 14 & -5 \\ 11 & -3 \end{vmatrix} = 13 \\ M_{33} &= \begin{vmatrix} 14 & 11 \\ 11 & 4 \end{vmatrix} = -65 \end{aligned}$$

   With the signs $(-1)^{i+j}$⁠, the cofactors of P are

   $$\begin{aligned} P_{11} &= -13,\quad P_{12} = 26,\quad P_{13} = -13 \\ P_{21} &= 26,\quad P_{22} = -39,\quad P_{23} = -13 \\ P_{31} &= -13,\quad P_{32} = -13,\quad P_{33} = -65 \end{aligned}$$

   This cofactor matrix is symmetric, so it equals its transpose, and every entry is $-13$ times the matching entry of A:

   $$\operatorname{adj}P = \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} = -13A.$$ $$\begin{aligned} [\operatorname{adj}A]^{-1} = P^{-1} &= \frac{1}{|P|}\operatorname{adj}P \\ &= \frac{1}{169}(-13A) = -\frac{1}{13}A. \end{aligned}$$
   For example, $M_{22} = 14(-1) - (-5)(-5)$ $= -14 - 25 = -39$ and $M_{33} = 14\cdot4 - 11\cdot11$ $= 56 - 121 = -65$⁠. Comparing with A: $-13\cdot1 = -13$⁠, $-13\cdot(-2) = 26$⁠, $-13\cdot5 = -65$⁠, and so on. Quick confirmation, using Theorem 1: $P\cdot\left(-\frac{1}{13}A\right)$ $= -\frac{1}{13}(\operatorname{adj}A)\,A$ $= -\frac{1}{13}(-13I) = I$ ✓.
   Section 4.4, Definition 2; Section 4.5, Theorem 4. Lesson: minors and cofactors

6. The adjoint of $A^{-1}$

   From Part (i), Step 3, $A^{-1} = -\frac{1}{13}P$⁠: every entry of $A^{-1}$ is $-\frac{1}{13}$ times the matching entry of P. A minor of $A^{-1}$ is a $2\times2$ determinant built from such entries, so it equals $\left(-\frac{1}{13}\right)^2 = \frac{1}{169}$ times the matching minor of P. For instance,

   $$\begin{aligned} M_{11}(A^{-1}) &= \begin{vmatrix} -\frac{4}{13} & \frac{3}{13} \\ \frac{3}{13} & \frac{1}{13} \end{vmatrix} = -\frac{4}{169} - \frac{9}{169} \\ &= -\frac{13}{169} = \frac{1}{169}(-13). \end{aligned}$$

   The signs $(-1)^{i+j}$ are the same, so every cofactor of $A^{-1}$ is $\frac{1}{169}$ times the cofactor of P in Part (i), Step 5, and therefore

   $$\begin{aligned} \operatorname{adj}(A^{-1}) &= \frac{1}{169}\operatorname{adj}P \\ &= \frac{1}{169}(-13A) = -\frac{1}{13}A. \end{aligned}$$

   Entry by entry, the cofactors of $A^{-1}$ are $-\frac{1}{13}$⁠, $\frac{2}{13}$⁠, $-\frac{1}{13}$ (row 1), $\frac{2}{13}$⁠, $-\frac{3}{13}$⁠, $-\frac{1}{13}$ (row 2) and $-\frac{1}{13}$⁠, $-\frac{1}{13}$⁠, $-\frac{5}{13}$ (row 3).

   This is the scaling rule of Basic 3 with $n = 2$⁠: multiplying every entry of a $2\times2$ determinant by $k$ multiplies its value by $k^2$⁠, because each of its two products $ad$ and $bc$ picks up $k$ twice. Here $k = -\frac{1}{13}$⁠, so the factor is $+\frac{1}{169}$⁠. It saves us from nine determinants full of fractions, and the one worked out above confirms it.
   Section 4.2.3, Remark (iii): if $A = kB$ of order $n$⁠, then $|A| = k^n|B|$⁠. Basics: scaling and adding matrices

7. Compare the two sides

   Part (i), Step 5 gave $[\operatorname{adj}A]^{-1} = -\frac{1}{13}A$ and Part (i), Step 6 gave $\operatorname{adj}(A^{-1}) = -\frac{1}{13}A$⁠. Both equal

   $$-\frac{1}{13}\begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix},$$

   so $[\operatorname{adj}A]^{-1} = \operatorname{adj}(A^{-1})$.

   "Verify" means compute both sides independently and see that they agree, which is what Part (i), Steps 5 and 6 did: one came from inverting adj A, the other from taking the adjoint of $A^{-1}$⁠. In words: inverting and taking the adjoint can be done in either order.
   Equal matrices: same order and equal corresponding entries. Basics: rows, columns and positions

Part (ii): $(A^{-1})^{-1} = A$

1. The determinant of $A^{-1}$

   $A^{-1} = -\frac{1}{13}P$ with P of order 3, so by the scaling rule with $n = 3$ and Part (i), Step 4:

   $$\begin{aligned} |A^{-1}| &= \left(-\frac{1}{13}\right)^3|P| = -\frac{1}{2197}\cdot169 \\ &= -\frac{1}{13}. \end{aligned}$$

   So $|A^{-1}| \ne 0$ and $A^{-1}$ is itself invertible.

   This matches $|A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$⁠, which follows from $|A|\,|A^{-1}| = |AA^{-1}| = |I| = 1$ (Theorem 3). Since $2197 = 13\cdot169$⁠, the fraction $\frac{169}{2197}$ reduces to $\frac{1}{13}$⁠.
   Section 4.2.3, Remark (iii); Section 4.5, Theorem 3: $|AB| = |A|\,|B|$⁠. Lesson: inverse of a matrix

2. Apply the inverse formula to $A^{-1}$

   Using $\operatorname{adj}(A^{-1}) = -\frac{1}{13}A$ from Part (i), Step 6:

   $$\begin{aligned} (A^{-1})^{-1} &= \frac{1}{|A^{-1}|}\operatorname{adj}(A^{-1}) \\ &= \frac{1}{-\frac{1}{13}}\left(-\frac{1}{13}A\right) \\ &= (-13)\left(-\frac{1}{13}\right)A = A. \end{aligned}$$

   So $(A^{-1})^{-1} = A$.

   The formula $M^{-1} = \frac{1}{|M|}\operatorname{adj}M$ holds for every non-singular M, and Part (ii), Step 1 showed $A^{-1}$ is non-singular. The result also makes sense directly: $AA^{-1} = A^{-1}A = I$ says that A undoes $A^{-1}$⁠, and an inverse is unique, so the inverse of $A^{-1}$ must be A.
   Section 4.5, Theorem 4. Basics: why the inverse divides by |A|

Every fact used, and where it comes from