Example 12: Find $\operatorname{adj}A$ for a $2\times2$ matrix

Basics you need

Solution

1. The order of $A$ and what we must build

   $A$ is $2\times2$⁠, so it is square and has an adjoint. By Definition 3 (Section 4.5.1), $\operatorname{adj}A$ is the transpose of the matrix of cofactors:

   $$\begin{aligned} \operatorname{adj}A &= \text{transpose of } \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}\\ &= \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix}. \end{aligned}$$

   So we need the four cofactors $A_{11}, A_{12}, A_{21}, A_{22}$⁠.

   Only square matrices have determinants, minors and cofactors, so only square matrices have adjoints. Writing the target pattern first shows where each cofactor will end up: $A_{21}$ goes to the top right and $A_{12}$ to the bottom left.
   Definition 3, Section 4.5.1. Lesson: the adjoint

2. The four minors

   Delete the row and column of each entry; one number remains:

   $$\begin{aligned} M_{11} &= 4 \quad (\text{delete row 1, column 1}),\\ M_{12} &= 1 \quad (\text{delete row 1, column 2}),\\ M_{21} &= 3 \quad (\text{delete row 2, column 1}),\\ M_{22} &= 2 \quad (\text{delete row 2, column 2}). \end{aligned}$$
   In a $2\times2$ matrix, deleting one row and one column leaves exactly the entry diagonally opposite. That is why $M_{11}$ is the bottom-right entry 4, and $M_{12}$ is the bottom-left entry 1.
   Definition 1, Section 4.4 (minor). Lesson: minors and cofactors

3. The four cofactors

   Attach the sign $(-1)^{i+j}$⁠:

   $$\begin{aligned} A_{11} &= (-1)^{1+1}(4) = 4,\\ A_{12} &= (-1)^{1+2}(1) = -1,\\ A_{21} &= (-1)^{2+1}(3) = -3,\\ A_{22} &= (-1)^{2+2}(2) = 2. \end{aligned}$$
   On a $2\times2$ grid the sign pattern is $+$ on the main diagonal and $-$ off it. So the two diagonal cofactors keep their sign and the two off-diagonal cofactors change sign. These match the textbook's values $A_{11} = 4$⁠, $A_{12} = -1$⁠, $A_{21} = -3$⁠, $A_{22} = 2$⁠.
   Definition 2, Section 4.4: $A_{ij} = (-1)^{i+j}M_{ij}$⁠. Basics: the + − + sign pattern

4. Matrix of cofactors, then transpose

   Put each cofactor where its element was:

   $$[A_{ij}] = \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}.$$

   Transpose it (row 1 becomes column 1, row 2 becomes column 2):

   $$\operatorname{adj}A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}.$$
   The definition says transpose of the cofactor matrix. Transposing moved $-1$ from position $(1,2)$ to $(2,1)$ and $-3$ from $(2,1)$ to $(1,2)$⁠; the diagonal entries 4 and 2 stay. Skipping this step is the most common error with adjoints (see Step 6 for what goes wrong).
   Transpose: $(i,j)$ entry of $P^{T}$ = $(j,i)$ entry of $P$⁠. Basics: transpose and identity

5. The $2\times2$ shortcut

   The textbook's Remark: for any square matrix of order 2,

   $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \ \text{ has } \ \operatorname{adj}A = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}.$$

   Here: swap $2$ and $4$ on the diagonal, and change the signs of $3$ and $1$⁠:

   $$\operatorname{adj}A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix},$$

   the same as Step 4.

   The shortcut works because each cofactor of a $2\times2$ entry is the diagonally opposite entry with the chessboard sign, so $A_{11} = a_{22}$⁠, $A_{12} = -a_{21}$⁠, $A_{21} = -a_{12}$⁠, $A_{22} = a_{11}$⁠, and the matrix of cofactors is $$[A_{ij}] = \begin{bmatrix} a_{22} & -a_{21} \\ -a_{12} & a_{11} \end{bmatrix}.$$ Transposing it swaps the two off-diagonal entries, which gives the Remark's matrix. The shortcut is only for order 2; a $3\times3$ adjoint needs all nine cofactors.
   Remark after Example 12, Section 4.5.1. Lesson: the adjoint

6. Check: $A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I$

   First the determinant: $|A| = 2\cdot 4 - 3\cdot 1 = 5$⁠. Now multiply, entry by entry. For $A(\operatorname{adj}A)$⁠:

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

   For $(\operatorname{adj}A)A$⁠:

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

   So both products are

   $$\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = |A|\,I,$$

   exactly as Theorem 1 predicts. If the transpose in Step 4 had been forgotten, the first product would be

   $$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -8 & 7 \end{bmatrix},$$

   which is not a multiple of $I$⁠: the check catches the mistake.

   The question asks only for $\operatorname{adj}A$⁠, so this step is a check, not part of the required answer. It is the best available one: Theorem 1 says the product must be $|A|$ on the diagonal and 0 elsewhere, and any wrong sign or a missing transpose breaks that pattern. (For the wrong matrix: $(1,1) = 2\cdot4 + 3\cdot(-3) = -1$⁠, $(1,2) = 2\cdot(-1) + 3\cdot2 = 4$⁠, $(2,1) = 1\cdot4 + 4\cdot(-3) = -8$⁠, $(2,2) = 1\cdot(-1) + 4\cdot2 = 7$⁠.)
   Theorem 1, Section 4.5: $A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I$⁠. Basics: why the inverse divides by |A|

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