Example 11: Minors and cofactors; check $a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}=0$

Basics you need

Solution

1. Name every element

   Reading row by row, the elements of $\Delta$ are

   $$\begin{aligned} &a_{11} = 2, \quad a_{12} = -3, \quad a_{13} = 5,\\ &a_{21} = 6, \quad a_{22} = 0, \quad a_{23} = 4,\\ &a_{31} = 1, \quad a_{32} = 5, \quad a_{33} = -7. \end{aligned}$$
   The question asks for "the minors and cofactors of the elements", which means all nine of them, and the final check uses $a_{11}, a_{12}, a_{13}$ by name. Writing the labels down once prevents the commonest slip in this topic: reading $a_{23}$ as "row 3, column 2".
   First subscript = row, second subscript = column. Basics: rows, columns and positions

2. The recipe and the sign pattern

   For each element $a_{ij}$⁠: delete row $i$ and column $j$⁠; the $2\times2$ determinant that remains is the minor $M_{ij}$ (Definition 1). Then the cofactor is $A_{ij} = (-1)^{i+j}M_{ij}$ (Definition 2). The factor $(-1)^{i+j}$ follows the chessboard pattern

   $$\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}$$

   So $A_{ij} = M_{ij}$ at the $+$ positions, and $A_{ij} = -M_{ij}$ at the $-$ positions $(1,2)$⁠, $(2,1)$⁠, $(2,3)$⁠, $(3,2)$⁠.

   A minor ignores where the element sits; the cofactor adds that information through the sign. The sign is exactly what makes "row times its own cofactors" equal to the determinant, as in the expansion formula of Section 4.2.3, which alternates $+,-,+$ along row 1.
   $(-1)^{\text{even}} = 1$ and $(-1)^{\text{odd}} = -1$⁠. For example $(-1)^{1+2} = (-1)^3 = -1$⁠. Basics: the + − + sign pattern

3. Row 1: $M_{11}, M_{12}, M_{13}$ and their cofactors

   $a_{11} = 2$⁠. Delete row 1 and column 1; the entries $0, 4, 5, -7$ remain:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = 0\cdot(-7) - 4\cdot 5 = -20,\\ A_{11} &= (-1)^{1+1}(-20) = -20. \end{aligned}$$

   $a_{12} = -3$⁠. Delete row 1 and column 2:

   $$\begin{aligned} M_{12} &= \begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = 6\cdot(-7) - 4\cdot 1 = -46,\\ A_{12} &= (-1)^{1+2}(-46) = 46. \end{aligned}$$

   $a_{13} = 5$⁠. Delete row 1 and column 3:

   $$\begin{aligned} M_{13} &= \begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix} = 6\cdot 5 - 0\cdot 1 = 30,\\ A_{13} &= (-1)^{1+3}(30) = 30. \end{aligned}$$
   The leftover entries keep their relative positions: what was upper left stays upper left. That matters, because swapping the two rows of a $2\times2$ determinant changes its sign. In $M_{12}$⁠, the sign $(-1)^3 = -1$ turns $-46$ into $+46$⁠.
   A $2\times2$ determinant is $ad - bc$ (Basic 2); keep brackets around negative numbers, $6\cdot(-7) = -42$⁠. Lesson: minors and cofactors

4. Row 2: $M_{21}, M_{22}, M_{23}$ and their cofactors

   $a_{21} = 6$⁠. Delete row 2 and column 1:

   $$\begin{aligned} M_{21} &= \begin{vmatrix} -3 & 5 \\ 5 & -7 \end{vmatrix} = (-3)(-7) - 5\cdot 5 = -4,\\ A_{21} &= (-1)^{2+1}(-4) = 4. \end{aligned}$$

   $a_{22} = 0$⁠. Delete row 2 and column 2:

   $$\begin{aligned} M_{22} &= \begin{vmatrix} 2 & 5 \\ 1 & -7 \end{vmatrix} = 2\cdot(-7) - 5\cdot 1 = -19,\\ A_{22} &= (-1)^{2+2}(-19) = -19. \end{aligned}$$

   $a_{23} = 4$⁠. Delete row 2 and column 3:

   $$\begin{aligned} M_{23} &= \begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = 2\cdot 5 - (-3)\cdot 1 = 13,\\ A_{23} &= (-1)^{2+3}(13) = -13. \end{aligned}$$
   The element $a_{22}$ is 0, but its minor and cofactor are not: a minor is built from the other entries, never from the element itself. The middle row has the pattern $-,+,-$⁠, so $A_{21}$ and $A_{23}$ change sign and $A_{22}$ does not.
   Subtracting a negative: $10 - (-3) = 10 + 3 = 13$⁠. Basics: algebra the chapter leans on

5. Row 3: $M_{31}, M_{32}, M_{33}$ and their cofactors

   $a_{31} = 1$⁠. Delete row 3 and column 1:

   $$\begin{aligned} M_{31} &= \begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = (-3)\cdot 4 - 5\cdot 0 = -12,\\ A_{31} &= (-1)^{3+1}(-12) = -12. \end{aligned}$$

   $a_{32} = 5$⁠. Delete row 3 and column 2 (see the figure below):

   $$\begin{aligned} M_{32} &= \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = 2\cdot 4 - 5\cdot 6 = -22,\\ A_{32} &= (-1)^{3+2}(-22) = 22. \end{aligned}$$

   $a_{33} = -7$⁠. Delete row 3 and column 3:

   $$\begin{aligned} M_{33} &= \begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = 2\cdot 0 - (-3)\cdot 6 = 18,\\ A_{33} &= (-1)^{3+3}(18) = 18. \end{aligned}$$
   Row 3 is the row the final check needs, so these three cofactors are the important ones: $A_{31} = -12$⁠, $A_{32} = 22$⁠, $A_{33} = 18$⁠. Notice that none of them uses an entry of row 3. Every one of them is built from rows 1 and 2 only. Step 8 depends on exactly this.
   Definition 1 (minor) and Definition 2 (cofactor), Section 4.4. Lesson: minors and cofactors

6. Collect all nine results

   Element	Minor $M_{ij}$	Sign	Cofactor $A_{ij}$
   $a_{11} = 2$	$-20$	$+$	$-20$
   $a_{12} = -3$	$-46$	$-$	$46$
   $a_{13} = 5$	$30$	$+$	$30$
   $a_{21} = 6$	$-4$	$-$	$4$
   $a_{22} = 0$	$-19$	$+$	$-19$
   $a_{23} = 4$	$13$	$-$	$-13$
   $a_{31} = 1$	$-12$	$+$	$-12$
   $a_{32} = 5$	$-22$	$-$	$22$
   $a_{33} = -7$	$18$	$+$	$18$

   A table lets you check the signs at a glance: in every row of the table with sign $-$⁠, the cofactor is the negative of the minor, and in every row with sign $+$ the two are equal.

7. Compute $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}$

   Take the elements of row 1, $a_{11} = 2$⁠, $a_{12} = -3$⁠, $a_{13} = 5$⁠, and the cofactors of row 3, $A_{31} = -12$⁠, $A_{32} = 22$⁠, $A_{33} = 18$⁠:

   $$\begin{aligned} &a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}\\ &= 2(-12) + (-3)(22) + 5(18)\\ &= -24 - 66 + 90 = 0. \end{aligned}$$

   So $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33} = 0$, as required.

   This is the "verify" part of the question: a direct substitution of the values from Steps 1 and 5. The elements come from one row and the cofactors from another; that mismatch is the whole point of the check.

8. Why the answer had to be 0

   The cofactors $A_{31}, A_{32}, A_{33}$ are built only from rows 1 and 2 (Step 5). So they are also the row-3 cofactors of the determinant $\Delta'$ obtained by replacing row 3 by a copy of row 1:

   $$\Delta' = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 2 & -3 & 5 \end{vmatrix}.$$

   Expanding $\Delta'$ along its row 3 multiplies the new row 3, which is $(2, -3, 5)$⁠, by those same cofactors. That is exactly the sum of Step 7:

   $$\Delta' = 2A_{31} + (-3)A_{32} + 5A_{33}.$$

   And $\Delta' = 0$⁠, because it has two equal rows. Expanding it instead along row 2 shows this: each minor of a row-2 element is built from rows 1 and 3 of $\Delta'$⁠, which are equal, so each minor has two equal rows and looks like

   $$\begin{vmatrix} p & q \\ p & q \end{vmatrix} = pq - qp = 0.$$

   So all three terms vanish and $\Delta' = 0$⁠. For the numbers here, expanding along row 1 gives the same:

   $$\begin{aligned} \Delta' &= 2(0\cdot5 - 4\cdot(-3)) + 3(6\cdot5 - 4\cdot2)\\ &\quad + 5(6\cdot(-3) - 0\cdot2)\\ &= 24 + 66 - 90 = 0. \end{aligned}$$
   This is the textbook's Note after Example 10: elements of one row multiplied by the cofactors of a different row always add up to 0. The check in this question is one instance of it. The Note itself only says "$= 0$ since $R_1$ and $R_2$ are identical"; this reprint has no section that proves "two equal rows give 0", so the row-2 argument above supplies that proof for any $3\times3$ determinant (it goes beyond the textbook's text). (The second term carries $+3$ because $-a_{12} = -(-3) = 3$ in the expansion along row 1.)
   Note after Example 10, Section 4.4; expansion along a row, Section 4.2.3. Lesson: expanding a 3 × 3 determinant

9. Contrast: a row with its own cofactors gives $\Delta$

   Row 1 with the cofactors of row 1:

   $$\begin{aligned} \Delta &= a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}\\ &= 2(-20) + (-3)(46) + 5(30)\\ &= -40 - 138 + 150 = -28. \end{aligned}$$

   Row 3 with the cofactors of row 3:

   $$\begin{aligned} \Delta &= a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33}\\ &= 1(-12) + 5(22) + (-7)(18)\\ &= -12 + 110 - 126 = -28. \end{aligned}$$

   Both give $\Delta = -28$⁠, as they must: a determinant can be expanded along any row.

   Matching a row with its own cofactors rebuilds $\Delta$⁠; matching it with another row's cofactors gives 0. Seeing both side by side is the best protection against mixing them up later, when this pair of facts becomes $A(\operatorname{adj}A) = |A|\,I$ in Section 4.5.
   Remark after Example 10: $\Delta$ = sum of the products of the elements of any row (or column) with their corresponding cofactors. Lesson: the adjoint

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