Exercise 4.3 · Q2: Minors and cofactors of $3\times3$ determinants

Basics you need

Part (i): the 3 × 3 identity determinant

1. (i) Minors of the first row

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1\cdot1 - 0\cdot0 = 1,\\ M_{12} &= \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0\cdot1 - 0\cdot0 = 0,\\ M_{13} &= \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0\cdot0 - 1\cdot0 = 0. \end{aligned}$$
   For $M_{11}$ delete $R_1$ and $C_1$⁠: what is left is the lower-right $2\times2$ block, $1, 0$ over $0, 1$⁠. For $M_{12}$ delete $R_1$ and $C_2$⁠: rows 2 and 3 keep their entries in columns 1 and 3, giving $0, 0$ over $0, 1$⁠. For $M_{13}$ delete $R_1$ and $C_3$⁠: $0, 1$ over $0, 0$⁠.
   Definition 1, Section 4.4. Lesson: minors and cofactors

2. (i) Minors of the second row

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0\cdot1 - 0\cdot0 = 0,\\ M_{22} &= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1\cdot1 - 0\cdot0 = 1,\\ M_{23} &= \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 1\cdot0 - 0\cdot0 = 0. \end{aligned}$$
   Each minor now uses rows 1 and 3. Deleting $C_2$ for $M_{22}$ leaves the corner entries $1, 0$ over $0, 1$⁠.

3. (i) Minors of the third row

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = 0\cdot0 - 0\cdot1 = 0,\\ M_{32} &= \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = 1\cdot0 - 0\cdot0 = 0,\\ M_{33} &= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1\cdot1 - 0\cdot0 = 1. \end{aligned}$$
   Each minor uses rows 1 and 2. The pattern is now clear: a diagonal element's minor is the $2\times2$ identity determinant, which is 1; every off-diagonal minor contains a row or a column of zeros, so it is 0.

4. (i) The cofactors

   Attach $(-1)^{i+j}$ to each minor:

   $$\begin{aligned} A_{11} &= (+1)(1) = 1,\\ A_{12} &= (-1)(0) = 0,\\ A_{13} &= (+1)(0) = 0,\\ A_{21} &= (-1)(0) = 0,\\ A_{22} &= (+1)(1) = 1,\\ A_{23} &= (-1)(0) = 0,\\ A_{31} &= (+1)(0) = 0,\\ A_{32} &= (-1)(0) = 0,\\ A_{33} &= (+1)(1) = 1. \end{aligned}$$
   A sign cannot change 0, and the three nonzero minors are on the diagonal, where $i + j$ is even and the sign is $+$⁠. So here every cofactor equals its minor.
   Definition 2, Section 4.4. Basics: the + − + sign pattern

5. (i) Check

   Row 1 with its cofactors: $1(1) + 0(0) + 0(0) = 1$⁠; likewise rows 2 and 3 give 1. So the identity determinant is 1, whichever row is used. Row 1 with the cofactors of row 2: $1(0) + 0(1) + 0(0) = 0$⁠. ✓

   Both checks of Basic 4 pass, so no sign or minor has been misplaced.

Part (ii): the determinant with rows (1, 0, 4), (3, 5, −1), (0, 1, 2)

1. (ii) Minors of the first row

   Delete $R_1$⁠; the minors use rows $(3, 5, -1)$ and $(0, 1, 2)$⁠:

   $$\begin{aligned} M_{11} &= \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix}\\ &= 5\cdot2 - (-1)\cdot1 = 11,\\ M_{12} &= \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = 3\cdot2 - (-1)\cdot0 = 6,\\ M_{13} &= \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3\cdot1 - 5\cdot0 = 3. \end{aligned}$$
   In $M_{11}$ the other-diagonal product is $(-1)\cdot1 = -1$⁠, and subtracting it adds 1: $10 + 1 = 11$⁠. In $M_{12}$ and $M_{13}$ the 0 in row 3 makes the second product 0.
   $2\times2$ rule (Basic 3). Basics: where the determinant comes from

2. (ii) Minors of the second row

   Delete $R_2$⁠; the minors use rows $(1, 0, 4)$ and $(0, 1, 2)$⁠:

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = 0\cdot2 - 4\cdot1 = -4,\\ M_{22} &= \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = 1\cdot2 - 4\cdot0 = 2,\\ M_{23} &= \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1\cdot1 - 0\cdot0 = 1. \end{aligned}$$
   For $M_{21}$⁠, deleting $C_1$ leaves columns 2 and 3: $0, 4$ over $1, 2$⁠. The main-diagonal product is 0, so the minor is $-4$⁠.

3. (ii) Minors of the third row

   Delete $R_3$⁠; the minors use rows $(1, 0, 4)$ and $(3, 5, -1)$⁠:

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix}\\ &= 0\cdot(-1) - 4\cdot5 = -20,\\ M_{32} &= \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix}\\ &= 1\cdot(-1) - 4\cdot3 = -13,\\ M_{33} &= \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = 1\cdot5 - 0\cdot3 = 5. \end{aligned}$$
   In $M_{32}$ both products matter: $-1 - 12 = -13$⁠. The $-1$ comes from the element $a_{23} = -1$⁠, which stays in the minor because we deleted column 2, not column 3.

4. (ii) The cofactors

   Attach the chessboard signs $+ - +\,/\,- + -\,/\,+ - +$⁠:

   $$\begin{aligned} A_{11} &= (+1)(11) = 11,\\ A_{12} &= (-1)(6) = -6,\\ A_{13} &= (+1)(3) = 3,\\ A_{21} &= (-1)(-4) = 4,\\ A_{22} &= (+1)(2) = 2,\\ A_{23} &= (-1)(1) = -1,\\ A_{31} &= (+1)(-20) = -20,\\ A_{32} &= (-1)(-13) = 13,\\ A_{33} &= (+1)(5) = 5. \end{aligned}$$
   The four positions with $i + j$ odd, namely (1, 2), (2, 1), (2, 3), (3, 2), change sign. Two of their minors are negative ($-4$ and $-13$⁠), so those cofactors become positive.
   Definition 2, Section 4.4. Basics: the + − + sign pattern

5. (ii) Check: every row gives the same determinant

   $$\begin{aligned} R_1:&\ \ 1(11) + 0(-6) + 4(3) = 23,\\ R_2:&\ \ 3(4) + 5(2) + (-1)(-1) = 23,\\ R_3:&\ \ 0(-20) + 1(13) + 2(5) = 23. \end{aligned}$$

   And row 1 with the cofactors of row 2: $1(4) + 0(2) + 4(-1) = 0$⁠. ✓

   The three expansions agree, so the determinant is 23, and the "wrong-row" sum is 0, as Section 4.4 says it must be. An error in any one minor or sign would almost certainly break at least one of these four equations.
   Section 4.4, Remark after Example 10 and the Note. Lesson: minors and cofactors

Every fact used, and where it comes from