Exercise 4.3 · Q1: Minors and cofactors of $2\times2$ determinants

Basics you need

Part (i): the determinant with rows (2, −4), (0, 3)

1. (i) Name the elements

   $$\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}:\qquad \begin{aligned} a_{11} &= 2, & a_{12} &= -4,\\ a_{21} &= 0, & a_{22} &= 3. \end{aligned}$$
   Minors and cofactors belong to positions, not to values, so we first fix which number sits where. The $-4$ is in row 1, column 2.
   Basics: rows, columns and positions

2. (i) The four minors

   Delete the row and column of each element; one element is left:

   Element	Delete	Left	Minor
   $a_{11} = 2$	$R_1$⁠, $C_1$	$a_{22}$	$M_{11} = 3$
   $a_{12} = -4$	$R_1$⁠, $C_2$	$a_{21}$	$M_{12} = 0$
   $a_{21} = 0$	$R_2$⁠, $C_1$	$a_{12}$	$M_{21} = -4$
   $a_{22} = 3$	$R_2$⁠, $C_2$	$a_{11}$	$M_{22} = 2$

   In a $2\times2$ determinant, deleting one row and one column removes three of the four elements, and the survivor is always the one diagonally opposite. The minor is the $1\times1$ determinant of that survivor, which is the survivor itself, sign included: $M_{21} = -4$⁠, not 4.
   Definition 1, Section 4.4; order-1 determinant, Section 4.2.1. Lesson: minors and cofactors

3. (i) The four cofactors

   $$\begin{aligned} A_{11} &= (-1)^{1+1}M_{11} = (+1)(3) = 3,\\ A_{12} &= (-1)^{1+2}M_{12} = (-1)(0) = 0,\\ A_{21} &= (-1)^{2+1}M_{21} = (-1)(-4) = 4,\\ A_{22} &= (-1)^{2+2}M_{22} = (+1)(2) = 2. \end{aligned}$$
   The exponent $i + j$ is even (2 or 4) on the main diagonal, so the sign is $+$ and the cofactor equals the minor. Off the diagonal $i + j = 3$ is odd, so the sign is $-$⁠: $A_{21}$ flips $-4$ to 4, and $A_{12}$ is $-0 = 0$⁠.
   Definition 2, Section 4.4: $A_{ij} = (-1)^{i+j}M_{ij}$⁠. Basics: the + − + sign pattern

4. (i) Check by expanding

   Directly, the determinant is $2\cdot3 - (-4)\cdot0 = 6$⁠. With cofactors:

   $$\begin{aligned} R_1:&\ \ a_{11}A_{11} + a_{12}A_{12}\\ &= 2(3) + (-4)(0) = 6,\\ R_2:&\ \ a_{21}A_{21} + a_{22}A_{22}\\ &= 0(4) + 3(2) = 6. \end{aligned}$$

   Both equal 6. ✓

   Every row (and column) expanded with its own cofactors must give the determinant. If a sign in Step 3 were wrong, one of these sums would come out different.
   Section 4.4, Remark after Example 10. Lesson: minors and cofactors

Part (ii): the determinant with rows (a, c), (b, d)

1. (ii) Name the elements

   $$\begin{vmatrix} a & c \\ b & d \end{vmatrix}:\qquad \begin{aligned} a_{11} &= a, & a_{12} &= c,\\ a_{21} &= b, & a_{22} &= d. \end{aligned}$$
   Note the arrangement: $c$ is in row 1 and $b$ is in row 2, not the other way round. Reading the letters in alphabetical order instead of by position is the easiest way to go wrong here.

2. (ii) The four minors

   Element	Delete	Left	Minor
   $a_{11} = a$	$R_1$⁠, $C_1$	$a_{22}$	$M_{11} = d$
   $a_{12} = c$	$R_1$⁠, $C_2$	$a_{21}$	$M_{12} = b$
   $a_{21} = b$	$R_2$⁠, $C_1$	$a_{12}$	$M_{21} = c$
   $a_{22} = d$	$R_2$⁠, $C_2$	$a_{11}$	$M_{22} = a$

   Same rule as Part (i): each minor is the element diagonally opposite.

3. (ii) The four cofactors

   $$\begin{aligned} A_{11} &= (-1)^{2}M_{11} = d,\\ A_{12} &= (-1)^{3}M_{12} = -b,\\ A_{21} &= (-1)^{3}M_{21} = -c,\\ A_{22} &= (-1)^{4}M_{22} = a. \end{aligned}$$
   The diagonal positions keep their minors; the off-diagonal positions change sign. With letters the sign cannot be absorbed into a number, so it stays visible as $-b$ and $-c$⁠.
   Definition 2, Section 4.4. Basics: the + − + sign pattern

4. (ii) Check by expanding

   Directly, the determinant is $ad - cb$⁠. With cofactors:

   $$\begin{aligned} R_1:&\ \ a(d) + c(-b) = ad - bc,\\ C_1:&\ \ a(d) + b(-c) = ad - bc. \end{aligned}$$

   Both give $ad - bc$⁠. ✓

   This works for every $a, b, c, d$⁠, so it confirms the general pattern: for a $2\times2$ determinant, the cofactors are "swap the diagonal, negate the off-diagonal". The same pattern reappears in Section 4.5 when the adjoint of a $2\times2$ matrix is formed.

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