Exercise 4.4 · Q2: Adjoint of a $3\times3$ matrix

Basics you need

Solution

1. Set up: order, entries and signs

   $A$ is square of order $3\times3$⁠, so $\operatorname{adj}A$ exists and is also $3\times3$⁠. We need nine cofactors. The sign $(-1)^{i+j}$ for each position is

   $$\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}$$
   Each cofactor is "minor with a sign", and the sign depends only on the position. Having the pattern in front of you avoids recomputing $(-1)^{i+j}$ nine times and catches the most common error, a wrong sign.
   Position $(1,1)$ has $i+j=2$⁠, even, so $+$⁠; moving one place along a row or column changes $i+j$ by 1 and flips the sign. Basics: the + − + sign pattern

2. Cofactors of the row 1 entries $1,\ -1,\ 2$

   For $a_{1j}$⁠, delete row 1 and column $j$⁠; the four entries left form $M_{1j}$⁠.

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3\cdot 1 - 5\cdot 0 \\ &= 3 - 0 = 3 \\ A_{11} &= (-1)^{1+1}\,(3) = 3 \\ M_{12} &= \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = 2\cdot 1 - 5\cdot (-2) \\ &= 2 + 10 = 12 \\ A_{12} &= (-1)^{1+2}\,(12) = -12 \\ M_{13} &= \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = 2\cdot 0 - 3\cdot (-2) \\ &= 0 + 6 = 6 \\ A_{13} &= (-1)^{1+3}\,(6) = 6 \end{aligned}$$
   Deleting row 1 always leaves rows 2 and 3, $(2,\ 3,\ 5)$ and $(-2,\ 0,\ 1)$⁠. Deleting column 1, 2 or 3 then keeps the other two columns. So $M_{11}$ uses columns 2, 3; $M_{12}$ uses columns 1, 3; $M_{13}$ uses columns 1, 2, each in their original order.
   Definitions 1 and 2, Section 4.4; the book works the same kind of calculation in Example 11. Lesson: minors and cofactors

3. Cofactors of the row 2 entries $2,\ 3,\ 5$

   Now delete row 2, which leaves rows 1 and 3, $(1,\ -1,\ 2)$ and $(-2,\ 0,\ 1)$⁠, and delete one column each time.

   $$\begin{aligned} M_{21} &= \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = (-1)\cdot 1 - 2\cdot 0 \\ &= -1 - 0 = -1 \\ A_{21} &= (-1)^{2+1}\,(-1) = 1 \\ M_{22} &= \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = 1\cdot 1 - 2\cdot (-2) \\ &= 1 + 4 = 5 \\ A_{22} &= (-1)^{2+2}\,(5) = 5 \\ M_{23} &= \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = 1\cdot 0 - (-1)\cdot (-2) \\ &= 0 - 2 = -2 \\ A_{23} &= (-1)^{2+3}\,(-2) = 2 \end{aligned}$$
   Row 2 is where the signs start with $-$⁠: positions $(2,1)$ and $(2,3)$ have odd $i+j$⁠. So $M_{21}=-1$ becomes $A_{21}=+1$⁠, and $M_{23}=-2$ becomes $A_{23}=+2$⁠. A negative minor in a minus position gives a positive cofactor.
   $(-1)\cdot(-1) = 1$⁠: two sign changes cancel. Basics: algebra the chapter leans on

4. Cofactors of the row 3 entries $-2,\ 0,\ 1$

   Delete row 3, which leaves rows 1 and 2, $(1,\ -1,\ 2)$ and $(2,\ 3,\ 5)$⁠.

   $$\begin{aligned} M_{31} &= \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = (-1)\cdot 5 - 2\cdot 3 \\ &= -5 - 6 = -11 \\ A_{31} &= (-1)^{3+1}\,(-11) = -11 \\ M_{32} &= \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = 1\cdot 5 - 2\cdot 2 \\ &= 5 - 4 = 1 \\ A_{32} &= (-1)^{3+2}\,(1) = -1 \\ M_{33} &= \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 1\cdot 3 - (-1)\cdot 2 \\ &= 3 + 2 = 5 \\ A_{33} &= (-1)^{3+3}\,(5) = 5 \end{aligned}$$
   The entry $a_{32}=0$ still has a cofactor. A cofactor depends on the other entries of the matrix, never on the entry itself, so a zero entry can have a nonzero cofactor (here $A_{32}=-1$⁠).

5. The matrix of cofactors

   Put each $A_{ij}$ in row $i$⁠, column $j$⁠:

   $$[A_{ij}] = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$$
   Row 1 holds $A_{11}, A_{12}, A_{13}$ from Step 2, row 2 holds the cofactors from Step 3, and row 3 those from Step 4. Keeping them in their own positions is what makes the next step a plain transpose.

6. Transpose to get $\operatorname{adj}A$

   Row 1 of $[A_{ij}]$⁠, $(3,\ -12,\ 6)$⁠, becomes column 1; row 2, $(1,\ 5,\ 2)$⁠, becomes column 2; row 3, $(-11,\ -1,\ 5)$⁠, becomes column 3:

   $$\begin{aligned} \operatorname{adj}A &= \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} \\ &= \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix} \end{aligned}$$
   This is Definition 3. The transpose is what makes Theorem 1 work: in $A(\operatorname{adj}A)$⁠, row $i$ of $A$ meets column $i$ of $\operatorname{adj}A$⁠, which holds the cofactors of row $i$⁠, so the diagonal entries are expansions of $|A|$⁠.
   The diagonal entries $3, 5, 5$ do not move under the transpose; every other cofactor swaps with its mirror image across the diagonal. Basics: transpose and identity

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