Example 13: Verify $A(\operatorname{adj}A)=|A|I$⁠; find $A^{-1}$

Basics you need

Solution

1. Compute $|A|$ by expanding along row 1

   No row or column of $A$ contains a zero, so no expansion is shorter than another; we use row 1, as the textbook does. The signs along row 1 are $+,-,+$⁠:

   $$|A| = 1\begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} - 3\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} + 3\begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix}.$$

   The three $2\times2$ determinants are

   $$\begin{aligned} \begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} &= 16 - 9 = 7,\\ \begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} &= 4 - 3 = 1,\\ \begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} &= 3 - 4 = -1. \end{aligned}$$

   So

   $$|A| = 1(7) - 3(1) + 3(-1) = 7 - 3 - 3 = 1.$$
   We need $|A|$ twice: for the right-hand side $|A|\,I$ of the identity, and to decide whether $A^{-1}$ exists. Here $|A| = 1 \ne 0$⁠, so $A$ is non-singular and invertible.
   Expansion along a row, Section 4.2.3; non-singular means $|A| \ne 0$ (Definition 5, Section 4.5). Lesson: expanding a 3 × 3 determinant

2. Cofactors of row 1

   The three minors of row 1 are the $2\times2$ determinants already found in Step 1:

   $$\begin{aligned} A_{11} &= (+1)\begin{vmatrix} 4 & 3 \\ 3 & 4 \end{vmatrix} = 7,\\ A_{12} &= (-1)\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = -1,\\ A_{13} &= (+1)\begin{vmatrix} 1 & 4 \\ 1 & 3 \end{vmatrix} = -1. \end{aligned}$$
   Expanding along row 1 is the same as $|A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$⁠, which is $1(7) + 3(-1) + 3(-1) = 1$⁠. The minus sign written in Step 1 in front of the middle term is the sign that turns $M_{12} = 1$ into $A_{12} = -1$⁠.
   $A_{ij} = (-1)^{i+j}M_{ij}$⁠, Definition 2, Section 4.4. Lesson: minors and cofactors

3. Cofactors of row 2

   Delete row 2 and, in turn, columns 1, 2, 3. The signs along row 2 are $-,+,-$⁠:

   $$\begin{aligned} A_{21} &= (-1)\begin{vmatrix} 3 & 3 \\ 3 & 4 \end{vmatrix} = -(12 - 9) = -3,\\ A_{22} &= (+1)\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 4 - 3 = 1,\\ A_{23} &= (-1)\begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = -(3 - 3) = 0. \end{aligned}$$
   $M_{23}$ is 0 because, after deleting row 2 and column 3, the two remaining rows are both $(1, 3)$⁠: rows 1 and 3 of $A$ agree in their first two entries. A $2\times2$ determinant with equal rows is $pq - qp = 0$⁠.
   Middle row of the sign pattern: $-,+,-$⁠. Basics: the + − + sign pattern

4. Cofactors of row 3

   Delete row 3 and, in turn, columns 1, 2, 3. The signs along row 3 are $+,-,+$⁠:

   $$\begin{aligned} A_{31} &= (+1)\begin{vmatrix} 3 & 3 \\ 4 & 3 \end{vmatrix} = 9 - 12 = -3,\\ A_{32} &= (-1)\begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = -(3 - 3) = 0,\\ A_{33} &= (+1)\begin{vmatrix} 1 & 3 \\ 1 & 4 \end{vmatrix} = 4 - 3 = 1. \end{aligned}$$
   Together with Steps 2 and 3 this gives all nine cofactors, matching the textbook: $A_{11} = 7$⁠, $A_{12} = -1$⁠, $A_{13} = -1$⁠, $A_{21} = -3$⁠, $A_{22} = 1$⁠, $A_{23} = 0$⁠, $A_{31} = -3$⁠, $A_{32} = 0$⁠, $A_{33} = 1$⁠. In $M_{31}$ the leftover entries are $3, 3$ over $4, 3$ (rows 1 and 2, columns 2 and 3), so it is $3\cdot3 - 3\cdot4 = -3$⁠.

5. Matrix of cofactors and $\operatorname{adj}A$

   Place each cofactor where its element sits:

   $$[A_{ij}] = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}.$$

   Transpose (row $i$ becomes column $i$⁠):

   $$\operatorname{adj}A = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}.$$
   The adjoint is the transpose of the cofactor matrix (Definition 3). So the cofactors of row 1, $7, -1, -1$⁠, form column 1 of $\operatorname{adj}A$⁠. This arrangement is what makes the product in Step 6 pair each row of $A$ with the cofactors of a row.
   Definition 3, Section 4.5.1. Lesson: the adjoint

6. Multiply $A(\operatorname{adj}A)$⁠, entry by entry

   Both matrices are $3\times3$⁠, so the product is defined and is $3\times3$⁠. Entry $(i,j)$ is row $i$ of $A$ times column $j$ of $\operatorname{adj}A$⁠:

   $$\begin{aligned} c_{11} &= 1\cdot7 + 3(-1) + 3(-1) = 1,\\ c_{12} &= 1(-3) + 3\cdot1 + 3\cdot0 = 0,\\ c_{13} &= 1(-3) + 3\cdot0 + 3\cdot1 = 0,\\ c_{21} &= 1\cdot7 + 4(-1) + 3(-1) = 0,\\ c_{22} &= 1(-3) + 4\cdot1 + 3\cdot0 = 1,\\ c_{23} &= 1(-3) + 4\cdot0 + 3\cdot1 = 0,\\ c_{31} &= 1\cdot7 + 3(-1) + 4(-1) = 0,\\ c_{32} &= 1(-3) + 3\cdot1 + 4\cdot0 = 0,\\ c_{33} &= 1(-3) + 3\cdot0 + 4\cdot1 = 1. \end{aligned}$$

   Assembling,

   $$A(\operatorname{adj}A) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = (1)\,I = |A|\,I.$$

   So $A(\operatorname{adj}A) = |A|\,I$ is verified.

   Look at what each entry is. $c_{22} = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$⁠: row 2 with its own cofactors, which is $|A| = 1$⁠. $c_{21} = a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}$⁠: row 2 with the cofactors of row 1, which is 0 (the Note after Example 10). Every diagonal entry is of the first kind and every off-diagonal entry of the second, which is why the result is $|A|$ times $I$⁠. Theorem 1 states this for every square matrix.
   Theorem 1, Section 4.5. Basics: matrix facts the chapter uses

7. Find $A^{-1}$

   Since $|A| = 1 \ne 0$⁠, $A$ is invertible (Theorem 4) and

   $$\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A\\ &= \frac{1}{1}\begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}. \end{aligned}$$
   Dividing $A(\operatorname{adj}A) = |A|\,I$ by the number $|A|$ gives $A\left(\frac{1}{|A|}\operatorname{adj}A\right) = I$⁠, and the same works for $(\operatorname{adj}A)A$⁠. A matrix that multiplies $A$ to $I$ on both sides is $A^{-1}$⁠. Here $|A| = 1$⁠, so the inverse is the adjoint itself. This step is impossible when $|A| = 0$⁠: there is nothing to divide by, and Theorem 4 says no inverse exists.
   Theorem 4, Section 4.5: $A$ is invertible exactly when $|A| \ne 0$⁠, and then $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$⁠. Lesson: inverse of a matrix

8. Check: $AA^{-1} = A^{-1}A = I$

   Because $A^{-1} = \operatorname{adj}A$ here, $AA^{-1}$ is exactly the product of Step 6, which is $I$⁠. For the other order, row $i$ of $A^{-1}$ times column $j$ of $A$⁠:

   $$\begin{aligned} d_{11} &= 7\cdot1 + (-3)\cdot1 + (-3)\cdot1 = 1,\\ d_{12} &= 7\cdot3 + (-3)\cdot4 + (-3)\cdot3 = 0,\\ d_{13} &= 7\cdot3 + (-3)\cdot3 + (-3)\cdot4 = 0,\\ d_{21} &= (-1)\cdot1 + 1\cdot1 + 0\cdot1 = 0,\\ d_{22} &= (-1)\cdot3 + 1\cdot4 + 0\cdot3 = 1,\\ d_{23} &= (-1)\cdot3 + 1\cdot3 + 0\cdot4 = 0,\\ d_{31} &= (-1)\cdot1 + 0\cdot1 + 1\cdot1 = 0,\\ d_{32} &= (-1)\cdot3 + 0\cdot4 + 1\cdot3 = 0,\\ d_{33} &= (-1)\cdot3 + 0\cdot3 + 1\cdot4 = 1. \end{aligned}$$

   So $A^{-1}A = I$ as well.

   An inverse must work on both sides: $AB = BA = I$ (Chapter 3, Section 3.7). Checking both orders confirms the answer completely. For square matrices one order actually implies the other (Example 19, Step 3 proves this with determinants), but the check costs little and catches arithmetic slips.
   Definition of the inverse, Chapter 3, Section 3.7. Basics: transpose and identity

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